YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z104-rotate.srs

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 64 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 10 ms)
↳6 QTRS
↳7 QTRSRRRProof (⇔, 15 ms)
↳8 QTRS
↳9 Overlay + Local Confluence (⇔, 24 ms)
↳10 QTRS
↳11 DependencyPairsProof (⇔, 26 ms)
↳12 QDP
↳13 DependencyGraphProof (⇔, 0 ms)
↳14 AND
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QReductionProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 MRRProof (⇔, 4 ms)
        ↳21 QDP
        ↳22 MRRProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QDPOrderProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPOrderProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPOrderProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 PisEmptyProof (⇔, 0 ms)
        ↳38 YES
    ↳39 QDP
        ↳40 UsableRulesProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QReductionProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 QDPOrderProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 PisEmptyProof (⇔, 0 ms)
        ↳47 YES
    ↳48 QDP
        ↳49 UsableRulesProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 QReductionProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 QDPOrderProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 PisEmptyProof (⇔, 0 ms)
        ↳56 YES
    ↳57 QDP
        ↳58 UsableRulesProof (⇔, 0 ms)
        ↳59 QDP
        ↳60 QReductionProof (⇔, 0 ms)
        ↳61 QDP
        ↳62 QDPSizeChangeProof (⇔, 0 ms)
        ↳63 YES
    ↳64 QDP
        ↳65 UsableRulesProof (⇔, 0 ms)
        ↳66 QDP
        ↳67 QReductionProof (⇔, 0 ms)
        ↳68 QDP
        ↳69 QDPSizeChangeProof (⇔, 0 ms)
        ↳70 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

begin(end(x)) → rewrite(end(x))
begin(c(x)) → rotate(cut(Cc(guess(x))))
begin(a(x)) → rotate(cut(Ca(guess(x))))
begin(d(x)) → rotate(cut(Cd(guess(x))))
begin(b(x)) → rotate(cut(Cb(guess(x))))
guess(c(x)) → Cc(guess(x))
guess(a(x)) → Ca(guess(x))
guess(d(x)) → Cd(guess(x))
guess(b(x)) → Cb(guess(x))
guess(c(x)) → moveleft(Bc(wait(x)))
guess(a(x)) → moveleft(Ba(wait(x)))
guess(d(x)) → moveleft(Bd(wait(x)))
guess(b(x)) → moveleft(Bb(wait(x)))
guess(end(x)) → finish(end(x))
Cc(moveleft(Bc(x))) → moveleft(Bc(Ac(x)))
Ca(moveleft(Bc(x))) → moveleft(Bc(Aa(x)))
Cd(moveleft(Bc(x))) → moveleft(Bc(Ad(x)))
Cb(moveleft(Bc(x))) → moveleft(Bc(Ab(x)))
Cc(moveleft(Ba(x))) → moveleft(Ba(Ac(x)))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cd(moveleft(Ba(x))) → moveleft(Ba(Ad(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Cc(moveleft(Bd(x))) → moveleft(Bd(Ac(x)))
Ca(moveleft(Bd(x))) → moveleft(Bd(Aa(x)))
Cd(moveleft(Bd(x))) → moveleft(Bd(Ad(x)))
Cb(moveleft(Bd(x))) → moveleft(Bd(Ab(x)))
Cc(moveleft(Bb(x))) → moveleft(Bb(Ac(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cd(moveleft(Bb(x))) → moveleft(Bb(Ad(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Bc(x))) → Dc(cut(goright(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bd(x))) → Dd(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Ac(x)) → Cc(goright(x))
goright(Aa(x)) → Ca(goright(x))
goright(Ad(x)) → Cd(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(c(x))) → moveleft(Bc(wait(x)))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(d(x))) → moveleft(Bd(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
goright(wait(end(x))) → finish(end(x))
Cc(finish(x)) → finish(c(x))
Ca(finish(x)) → finish(a(x))
Cd(finish(x)) → finish(d(x))
Cb(finish(x)) → finish(b(x))
cut(finish(x)) → finish2(x)
Dc(finish2(x)) → finish2(c(x))
Da(finish2(x)) → finish2(a(x))
Dd(finish2(x)) → finish2(d(x))
Db(finish2(x)) → finish2(b(x))
rotate(finish2(x)) → rewrite(x)
rewrite(c(c(c(a(x))))) → begin(d(d(x)))
rewrite(d(b(x))) → begin(c(c(x)))
rewrite(c(x)) → begin(a(a(a(a(x)))))
rewrite(d(x)) → begin(b(b(b(b(x)))))
rewrite(b(d(x))) → begin(c(c(x)))
rewrite(a(c(c(c(x))))) → begin(d(d(x)))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

end(begin(x)) → end(rewrite(x))
c(begin(x)) → guess(Cc(cut(rotate(x))))
a(begin(x)) → guess(Ca(cut(rotate(x))))
d(begin(x)) → guess(Cd(cut(rotate(x))))
b(begin(x)) → guess(Cb(cut(rotate(x))))
c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
c(guess(x)) → wait(Bc(moveleft(x)))
a(guess(x)) → wait(Ba(moveleft(x)))
d(guess(x)) → wait(Bd(moveleft(x)))
b(guess(x)) → wait(Bb(moveleft(x)))
end(guess(x)) → end(finish(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
end(wait(goright(x))) → end(finish(x))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish(cut(x)) → finish2(x)
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))
finish2(rotate(x)) → rewrite(x)
a(c(c(c(rewrite(x))))) → d(d(begin(x)))
b(d(rewrite(x))) → c(c(begin(x)))
c(rewrite(x)) → a(a(a(a(begin(x)))))
d(rewrite(x)) → b(b(b(b(begin(x)))))
d(b(rewrite(x))) → c(c(begin(x)))
c(c(c(a(rewrite(x))))) → d(d(begin(x)))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 83 + x1   
POL(Ab(x1)) = 135 + x1   
POL(Ac(x1)) = 338 + x1   
POL(Ad(x1)) = 546 + x1   
POL(Ba(x1)) = 83 + x1   
POL(Bb(x1)) = 135 + x1   
POL(Bc(x1)) = 338 + x1   
POL(Bd(x1)) = 546 + x1   
POL(Ca(x1)) = 83 + x1   
POL(Cb(x1)) = 135 + x1   
POL(Cc(x1)) = 338 + x1   
POL(Cd(x1)) = 546 + x1   
POL(Da(x1)) = 83 + x1   
POL(Db(x1)) = 135 + x1   
POL(Dc(x1)) = 338 + x1   
POL(Dd(x1)) = 546 + x1   
POL(a(x1)) = 83 + x1   
POL(b(x1)) = 135 + x1   
POL(begin(x1)) = 5 + x1   
POL(c(x1)) = 338 + x1   
POL(cut(x1)) = 2 + x1   
POL(d(x1)) = 546 + x1   
POL(end(x1)) = x1   
POL(finish(x1)) = x1   
POL(finish2(x1)) = 1 + x1   
POL(goright(x1)) = x1   
POL(guess(x1)) = 2 + x1   
POL(moveleft(x1)) = x1   
POL(rewrite(x1)) = x1   
POL(rotate(x1)) = x1   
POL(wait(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

end(begin(x)) → end(rewrite(x))
c(begin(x)) → guess(Cc(cut(rotate(x))))
a(begin(x)) → guess(Ca(cut(rotate(x))))
d(begin(x)) → guess(Cd(cut(rotate(x))))
b(begin(x)) → guess(Cb(cut(rotate(x))))
c(guess(x)) → wait(Bc(moveleft(x)))
a(guess(x)) → wait(Ba(moveleft(x)))
d(guess(x)) → wait(Bd(moveleft(x)))
b(guess(x)) → wait(Bb(moveleft(x)))
end(guess(x)) → end(finish(x))
end(wait(goright(x))) → end(finish(x))
finish(cut(x)) → finish2(x)
finish2(rotate(x)) → rewrite(x)
c(rewrite(x)) → a(a(a(a(begin(x)))))
d(rewrite(x)) → b(b(b(b(begin(x)))))


(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))
a(c(c(c(rewrite(x))))) → d(d(begin(x)))
b(d(rewrite(x))) → c(c(begin(x)))
d(b(rewrite(x))) → c(c(begin(x)))
c(c(c(a(rewrite(x))))) → d(d(begin(x)))

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = 1 + x1   
POL(Ac(x1)) = x1   
POL(Ad(x1)) = x1   
POL(Ba(x1)) = x1   
POL(Bb(x1)) = 1 + x1   
POL(Bc(x1)) = x1   
POL(Bd(x1)) = x1   
POL(Ca(x1)) = x1   
POL(Cb(x1)) = 1 + x1   
POL(Cc(x1)) = x1   
POL(Cd(x1)) = x1   
POL(Da(x1)) = x1   
POL(Db(x1)) = 1 + x1   
POL(Dc(x1)) = x1   
POL(Dd(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(begin(x1)) = x1   
POL(c(x1)) = x1   
POL(cut(x1)) = x1   
POL(d(x1)) = x1   
POL(finish(x1)) = x1   
POL(finish2(x1)) = x1   
POL(goright(x1)) = x1   
POL(guess(x1)) = x1   
POL(moveleft(x1)) = x1   
POL(rewrite(x1)) = x1   
POL(wait(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

b(d(rewrite(x))) → c(c(begin(x)))
d(b(rewrite(x))) → c(c(begin(x)))


(6) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))
a(c(c(c(rewrite(x))))) → d(d(begin(x)))
c(c(c(a(rewrite(x))))) → d(d(begin(x)))

Q is empty.

(7) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Ac(x1)) = x1   
POL(Ad(x1)) = x1   
POL(Ba(x1)) = x1   
POL(Bb(x1)) = x1   
POL(Bc(x1)) = x1   
POL(Bd(x1)) = x1   
POL(Ca(x1)) = x1   
POL(Cb(x1)) = x1   
POL(Cc(x1)) = x1   
POL(Cd(x1)) = x1   
POL(Da(x1)) = x1   
POL(Db(x1)) = x1   
POL(Dc(x1)) = x1   
POL(Dd(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(begin(x1)) = x1   
POL(c(x1)) = x1   
POL(cut(x1)) = x1   
POL(d(x1)) = x1   
POL(finish(x1)) = x1   
POL(finish2(x1)) = x1   
POL(goright(x1)) = x1   
POL(guess(x1)) = x1   
POL(moveleft(x1)) = x1   
POL(rewrite(x1)) = 1 + x1   
POL(wait(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a(c(c(c(rewrite(x))))) → d(d(begin(x)))
c(c(c(a(rewrite(x))))) → d(d(begin(x)))


(8) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

Q is empty.

(9) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(10) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(11) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BC(moveleft(Cc(x))) → AC(Bc(moveleft(x)))
BC(moveleft(Cc(x))) → BC(moveleft(x))
BC(moveleft(Ca(x))) → AA(Bc(moveleft(x)))
BC(moveleft(Ca(x))) → BC(moveleft(x))
BC(moveleft(Cd(x))) → AD(Bc(moveleft(x)))
BC(moveleft(Cd(x))) → BC(moveleft(x))
BC(moveleft(Cb(x))) → AB(Bc(moveleft(x)))
BC(moveleft(Cb(x))) → BC(moveleft(x))
BA(moveleft(Cc(x))) → AC(Ba(moveleft(x)))
BA(moveleft(Cc(x))) → BA(moveleft(x))
BA(moveleft(Ca(x))) → AA(Ba(moveleft(x)))
BA(moveleft(Ca(x))) → BA(moveleft(x))
BA(moveleft(Cd(x))) → AD(Ba(moveleft(x)))
BA(moveleft(Cd(x))) → BA(moveleft(x))
BA(moveleft(Cb(x))) → AB(Ba(moveleft(x)))
BA(moveleft(Cb(x))) → BA(moveleft(x))
BD(moveleft(Cc(x))) → AC(Bd(moveleft(x)))
BD(moveleft(Cc(x))) → BD(moveleft(x))
BD(moveleft(Ca(x))) → AA(Bd(moveleft(x)))
BD(moveleft(Ca(x))) → BD(moveleft(x))
BD(moveleft(Cd(x))) → AD(Bd(moveleft(x)))
BD(moveleft(Cd(x))) → BD(moveleft(x))
BD(moveleft(Cb(x))) → AB(Bd(moveleft(x)))
BD(moveleft(Cb(x))) → BD(moveleft(x))
BB(moveleft(Cc(x))) → AC(Bb(moveleft(x)))
BB(moveleft(Cc(x))) → BB(moveleft(x))
BB(moveleft(Ca(x))) → AA(Bb(moveleft(x)))
BB(moveleft(Ca(x))) → BB(moveleft(x))
BB(moveleft(Cd(x))) → AD(Bb(moveleft(x)))
BB(moveleft(Cd(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → AB(Bb(moveleft(x)))
BB(moveleft(Cb(x))) → BB(moveleft(x))
C(wait(goright(x))) → BC(moveleft(x))
A(wait(goright(x))) → BA(moveleft(x))
D(wait(goright(x))) → BD(moveleft(x))
B(wait(goright(x))) → BB(moveleft(x))
FINISH(Cc(x)) → C(finish(x))
FINISH(Cc(x)) → FINISH(x)
FINISH(Ca(x)) → A(finish(x))
FINISH(Ca(x)) → FINISH(x)
FINISH(Cd(x)) → D(finish(x))
FINISH(Cd(x)) → FINISH(x)
FINISH(Cb(x)) → B(finish(x))
FINISH(Cb(x)) → FINISH(x)
FINISH2(Dc(x)) → C(finish2(x))
FINISH2(Dc(x)) → FINISH2(x)
FINISH2(Da(x)) → A(finish2(x))
FINISH2(Da(x)) → FINISH2(x)
FINISH2(Dd(x)) → D(finish2(x))
FINISH2(Dd(x)) → FINISH2(x)
FINISH2(Db(x)) → B(finish2(x))
FINISH2(Db(x)) → FINISH2(x)

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 28 less nodes.

(14) Complex Obligation (AND)

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BB(moveleft(Ca(x))) → BB(moveleft(x))
BB(moveleft(Cc(x))) → BB(moveleft(x))
BB(moveleft(Cd(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → BB(moveleft(x))

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(16) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BB(moveleft(Ca(x))) → BB(moveleft(x))
BB(moveleft(Cc(x))) → BB(moveleft(x))
BB(moveleft(Cd(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → BB(moveleft(x))

R is empty.
The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(18) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(19) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BB(moveleft(Ca(x))) → BB(moveleft(x))
BB(moveleft(Cc(x))) → BB(moveleft(x))
BB(moveleft(Cd(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → BB(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

BB(moveleft(Ca(x))) → BB(moveleft(x))


Used ordering: Polynomial interpretation [POLO]:

POL(BB(x1)) = 2·x1   
POL(Ca(x1)) = 1 + x1   
POL(Cb(x1)) = 2·x1   
POL(Cc(x1)) = 2·x1   
POL(Cd(x1)) = 2·x1   
POL(moveleft(x1)) = 2·x1   

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BB(moveleft(Cc(x))) → BB(moveleft(x))
BB(moveleft(Cd(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → BB(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

BB(moveleft(Cc(x))) → BB(moveleft(x))


Used ordering: Polynomial interpretation [POLO]:

POL(BB(x1)) = 2·x1   
POL(Cb(x1)) = 2·x1   
POL(Cc(x1)) = 1 + x1   
POL(Cd(x1)) = 2·x1   
POL(moveleft(x1)) = x1   

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BB(moveleft(Cd(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → BB(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BB(moveleft(Cb(x))) → BB(moveleft(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(BB(x1)) = 4·x1   
POL(Cb(x1)) = 5 + 2·x1   
POL(Cd(x1)) = 2·x1   
POL(moveleft(x1)) = 5·x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BB(moveleft(Cd(x))) → BB(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BB(moveleft(Cd(x))) → BB(moveleft(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(BB(x1)) = x1   
POL(Cd(x1)) = 1 + x1   
POL(moveleft(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(27) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) YES

(30) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BD(moveleft(Ca(x))) → BD(moveleft(x))
BD(moveleft(Cc(x))) → BD(moveleft(x))
BD(moveleft(Cd(x))) → BD(moveleft(x))
BD(moveleft(Cb(x))) → BD(moveleft(x))

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(31) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BD(moveleft(Ca(x))) → BD(moveleft(x))
BD(moveleft(Cc(x))) → BD(moveleft(x))
BD(moveleft(Cd(x))) → BD(moveleft(x))
BD(moveleft(Cb(x))) → BD(moveleft(x))

R is empty.
The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(33) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BD(moveleft(Ca(x))) → BD(moveleft(x))
BD(moveleft(Cc(x))) → BD(moveleft(x))
BD(moveleft(Cd(x))) → BD(moveleft(x))
BD(moveleft(Cb(x))) → BD(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BD(moveleft(Ca(x))) → BD(moveleft(x))
BD(moveleft(Cc(x))) → BD(moveleft(x))
BD(moveleft(Cd(x))) → BD(moveleft(x))
BD(moveleft(Cb(x))) → BD(moveleft(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( BD(x1) ) = max{0, 2x1 - 1}

POL( moveleft(x1) ) = 2x1

POL( Ca(x1) ) = 2x1 + 2

POL( Cc(x1) ) = 2x1 + 1

POL( Cd(x1) ) = 2x1 + 2

POL( Cb(x1) ) = 2x1 + 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(36) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(38) YES

(39) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BA(moveleft(Ca(x))) → BA(moveleft(x))
BA(moveleft(Cc(x))) → BA(moveleft(x))
BA(moveleft(Cd(x))) → BA(moveleft(x))
BA(moveleft(Cb(x))) → BA(moveleft(x))

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(40) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BA(moveleft(Ca(x))) → BA(moveleft(x))
BA(moveleft(Cc(x))) → BA(moveleft(x))
BA(moveleft(Cd(x))) → BA(moveleft(x))
BA(moveleft(Cb(x))) → BA(moveleft(x))

R is empty.
The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(42) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BA(moveleft(Ca(x))) → BA(moveleft(x))
BA(moveleft(Cc(x))) → BA(moveleft(x))
BA(moveleft(Cd(x))) → BA(moveleft(x))
BA(moveleft(Cb(x))) → BA(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BA(moveleft(Ca(x))) → BA(moveleft(x))
BA(moveleft(Cc(x))) → BA(moveleft(x))
BA(moveleft(Cd(x))) → BA(moveleft(x))
BA(moveleft(Cb(x))) → BA(moveleft(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(BA(x1)) = x1   
POL(Ca(x1)) = 1 + x1   
POL(Cb(x1)) = 1 + x1   
POL(Cc(x1)) = 1 + x1   
POL(Cd(x1)) = 1 + x1   
POL(moveleft(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(45) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(46) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(47) YES

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BC(moveleft(Ca(x))) → BC(moveleft(x))
BC(moveleft(Cc(x))) → BC(moveleft(x))
BC(moveleft(Cd(x))) → BC(moveleft(x))
BC(moveleft(Cb(x))) → BC(moveleft(x))

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(49) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(50) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BC(moveleft(Ca(x))) → BC(moveleft(x))
BC(moveleft(Cc(x))) → BC(moveleft(x))
BC(moveleft(Cd(x))) → BC(moveleft(x))
BC(moveleft(Cb(x))) → BC(moveleft(x))

R is empty.
The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(51) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(52) Obligation:

Q DP problem:
The TRS P consists of the following rules:

BC(moveleft(Ca(x))) → BC(moveleft(x))
BC(moveleft(Cc(x))) → BC(moveleft(x))
BC(moveleft(Cd(x))) → BC(moveleft(x))
BC(moveleft(Cb(x))) → BC(moveleft(x))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


BC(moveleft(Ca(x))) → BC(moveleft(x))
BC(moveleft(Cc(x))) → BC(moveleft(x))
BC(moveleft(Cd(x))) → BC(moveleft(x))
BC(moveleft(Cb(x))) → BC(moveleft(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( BC(x1) ) = max{0, 2x1 - 1}

POL( moveleft(x1) ) = 2x1

POL( Ca(x1) ) = 2x1 + 2

POL( Cc(x1) ) = 2x1 + 1

POL( Cd(x1) ) = 2x1 + 2

POL( Cb(x1) ) = 2x1 + 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(54) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(55) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(56) YES

(57) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FINISH2(Da(x)) → FINISH2(x)
FINISH2(Dc(x)) → FINISH2(x)
FINISH2(Dd(x)) → FINISH2(x)
FINISH2(Db(x)) → FINISH2(x)

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(58) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(59) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FINISH2(Da(x)) → FINISH2(x)
FINISH2(Dc(x)) → FINISH2(x)
FINISH2(Dd(x)) → FINISH2(x)
FINISH2(Db(x)) → FINISH2(x)

R is empty.
The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(60) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(61) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FINISH2(Da(x)) → FINISH2(x)
FINISH2(Dc(x)) → FINISH2(x)
FINISH2(Dd(x)) → FINISH2(x)
FINISH2(Db(x)) → FINISH2(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(62) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • FINISH2(Da(x)) → FINISH2(x)
    The graph contains the following edges 1 > 1

  • FINISH2(Dc(x)) → FINISH2(x)
    The graph contains the following edges 1 > 1

  • FINISH2(Dd(x)) → FINISH2(x)
    The graph contains the following edges 1 > 1

  • FINISH2(Db(x)) → FINISH2(x)
    The graph contains the following edges 1 > 1

(63) YES

(64) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FINISH(Ca(x)) → FINISH(x)
FINISH(Cc(x)) → FINISH(x)
FINISH(Cd(x)) → FINISH(x)
FINISH(Cb(x)) → FINISH(x)

The TRS R consists of the following rules:

c(guess(x)) → guess(Cc(x))
a(guess(x)) → guess(Ca(x))
d(guess(x)) → guess(Cd(x))
b(guess(x)) → guess(Cb(x))
Bc(moveleft(Cc(x))) → Ac(Bc(moveleft(x)))
Bc(moveleft(Ca(x))) → Aa(Bc(moveleft(x)))
Bc(moveleft(Cd(x))) → Ad(Bc(moveleft(x)))
Bc(moveleft(Cb(x))) → Ab(Bc(moveleft(x)))
Ba(moveleft(Cc(x))) → Ac(Ba(moveleft(x)))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cd(x))) → Ad(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bd(moveleft(Cc(x))) → Ac(Bd(moveleft(x)))
Bd(moveleft(Ca(x))) → Aa(Bd(moveleft(x)))
Bd(moveleft(Cd(x))) → Ad(Bd(moveleft(x)))
Bd(moveleft(Cb(x))) → Ab(Bd(moveleft(x)))
Bb(moveleft(Cc(x))) → Ac(Bb(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cd(x))) → Ad(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Bc(moveleft(cut(x))) → goright(cut(Dc(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bd(moveleft(cut(x))) → goright(cut(Dd(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Ac(goright(x)) → goright(Cc(x))
Aa(goright(x)) → goright(Ca(x))
Ad(goright(x)) → goright(Cd(x))
Ab(goright(x)) → goright(Cb(x))
c(wait(goright(x))) → wait(Bc(moveleft(x)))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
d(wait(goright(x))) → wait(Bd(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Cc(x)) → c(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cd(x)) → d(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Dc(x)) → c(finish2(x))
finish2(Da(x)) → a(finish2(x))
finish2(Dd(x)) → d(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(65) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(66) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FINISH(Ca(x)) → FINISH(x)
FINISH(Cc(x)) → FINISH(x)
FINISH(Cd(x)) → FINISH(x)
FINISH(Cb(x)) → FINISH(x)

R is empty.
The set Q consists of the following terms:

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

We have to consider all minimal (P,Q,R)-chains.

(67) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

c(guess(x0))
a(guess(x0))
d(guess(x0))
b(guess(x0))
Bc(moveleft(Cc(x0)))
Bc(moveleft(Ca(x0)))
Bc(moveleft(Cd(x0)))
Bc(moveleft(Cb(x0)))
Ba(moveleft(Cc(x0)))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cd(x0)))
Ba(moveleft(Cb(x0)))
Bd(moveleft(Cc(x0)))
Bd(moveleft(Ca(x0)))
Bd(moveleft(Cd(x0)))
Bd(moveleft(Cb(x0)))
Bb(moveleft(Cc(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cd(x0)))
Bb(moveleft(Cb(x0)))
Bc(moveleft(cut(x0)))
Ba(moveleft(cut(x0)))
Bd(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Ac(goright(x0))
Aa(goright(x0))
Ad(goright(x0))
Ab(goright(x0))
c(wait(goright(x0)))
a(wait(goright(x0)))
d(wait(goright(x0)))
b(wait(goright(x0)))
finish(Cc(x0))
finish(Ca(x0))
finish(Cd(x0))
finish(Cb(x0))
finish2(Dc(x0))
finish2(Da(x0))
finish2(Dd(x0))
finish2(Db(x0))

(68) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FINISH(Ca(x)) → FINISH(x)
FINISH(Cc(x)) → FINISH(x)
FINISH(Cd(x)) → FINISH(x)
FINISH(Cb(x)) → FINISH(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(69) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • FINISH(Ca(x)) → FINISH(x)
    The graph contains the following edges 1 > 1

  • FINISH(Cc(x)) → FINISH(x)
    The graph contains the following edges 1 > 1

  • FINISH(Cd(x)) → FINISH(x)
    The graph contains the following edges 1 > 1

  • FINISH(Cb(x)) → FINISH(x)
    The graph contains the following edges 1 > 1

(70) YES