YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z101-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 (⇔, 42 ms)
↳4 QTRS
↳5 Overlay + Local Confluence (⇔, 0 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 0 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 AND
    ↳11 QDP
        ↳12 UsableRulesProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QReductionProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 MRRProof (⇔, 5 ms)
        ↳17 QDP
        ↳18 QDPOrderProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 PisEmptyProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 UsableRulesProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QReductionProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 MRRProof (⇔, 0 ms)
        ↳28 QDP
        ↳29 QDPOrderProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 PisEmptyProof (⇔, 0 ms)
        ↳32 YES
    ↳33 QDP
        ↳34 UsableRulesProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 QReductionProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QDPSizeChangeProof (⇔, 0 ms)
        ↳39 YES
    ↳40 QDP
        ↳41 UsableRulesProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 QReductionProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 QDPSizeChangeProof (⇔, 0 ms)
        ↳46 YES

(0) Obligation:

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

begin(end(x)) → rewrite(end(x))
begin(a(x)) → rotate(cut(Ca(guess(x))))
begin(b(x)) → rotate(cut(Cb(guess(x))))
guess(a(x)) → Ca(guess(x))
guess(b(x)) → Cb(guess(x))
guess(a(x)) → moveleft(Ba(wait(x)))
guess(b(x)) → moveleft(Bb(wait(x)))
guess(end(x)) → finish(end(x))
Ca(moveleft(Ba(x))) → moveleft(Ba(Aa(x)))
Cb(moveleft(Ba(x))) → moveleft(Ba(Ab(x)))
Ca(moveleft(Bb(x))) → moveleft(Bb(Aa(x)))
Cb(moveleft(Bb(x))) → moveleft(Bb(Ab(x)))
cut(moveleft(Ba(x))) → Da(cut(goright(x)))
cut(moveleft(Bb(x))) → Db(cut(goright(x)))
goright(Aa(x)) → Ca(goright(x))
goright(Ab(x)) → Cb(goright(x))
goright(wait(a(x))) → moveleft(Ba(wait(x)))
goright(wait(b(x))) → moveleft(Bb(wait(x)))
goright(wait(end(x))) → finish(end(x))
Ca(finish(x)) → finish(a(x))
Cb(finish(x)) → finish(b(x))
cut(finish(x)) → finish2(x)
Da(finish2(x)) → finish2(a(x))
Db(finish2(x)) → finish2(b(x))
rotate(finish2(x)) → rewrite(x)
rewrite(a(a(x))) → begin(b(b(b(x))))
rewrite(b(b(b(b(b(x)))))) → begin(a(a(a(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))
a(begin(x)) → guess(Ca(cut(rotate(x))))
b(begin(x)) → guess(Cb(cut(rotate(x))))
a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
a(guess(x)) → wait(Ba(moveleft(x)))
b(guess(x)) → wait(Bb(moveleft(x)))
end(guess(x)) → end(finish(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
end(wait(goright(x))) → end(finish(x))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish(cut(x)) → finish2(x)
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))
finish2(rotate(x)) → rewrite(x)
a(a(rewrite(x))) → b(b(b(begin(x))))
b(b(b(b(b(rewrite(x)))))) → a(a(a(begin(x))))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 48 + x1   
POL(Ab(x1)) = 30 + x1   
POL(Ba(x1)) = 48 + x1   
POL(Bb(x1)) = 30 + x1   
POL(Ca(x1)) = 48 + x1   
POL(Cb(x1)) = 30 + x1   
POL(Da(x1)) = 48 + x1   
POL(Db(x1)) = 30 + x1   
POL(a(x1)) = 48 + x1   
POL(b(x1)) = 30 + x1   
POL(begin(x1)) = 5 + x1   
POL(cut(x1)) = 2 + 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))
a(begin(x)) → guess(Ca(cut(rotate(x))))
b(begin(x)) → guess(Cb(cut(rotate(x))))
a(guess(x)) → wait(Ba(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)
a(a(rewrite(x))) → b(b(b(begin(x))))
b(b(b(b(b(rewrite(x)))))) → a(a(a(begin(x))))


(4) Obligation:

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

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

Q is empty.

(5) Overlay + Local Confluence (EQUIVALENT transformation)

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

(6) Obligation:

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

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

BA(moveleft(Ca(x))) → AA(Ba(moveleft(x)))
BA(moveleft(Ca(x))) → BA(moveleft(x))
BA(moveleft(Cb(x))) → AB(Ba(moveleft(x)))
BA(moveleft(Cb(x))) → BA(moveleft(x))
BB(moveleft(Ca(x))) → AA(Bb(moveleft(x)))
BB(moveleft(Ca(x))) → BB(moveleft(x))
BB(moveleft(Cb(x))) → AB(Bb(moveleft(x)))
BB(moveleft(Cb(x))) → BB(moveleft(x))
A(wait(goright(x))) → BA(moveleft(x))
B(wait(goright(x))) → BB(moveleft(x))
FINISH(Ca(x)) → A(finish(x))
FINISH(Ca(x)) → FINISH(x)
FINISH(Cb(x)) → B(finish(x))
FINISH(Cb(x)) → FINISH(x)
FINISH2(Da(x)) → A(finish2(x))
FINISH2(Da(x)) → FINISH2(x)
FINISH2(Db(x)) → B(finish2(x))
FINISH2(Db(x)) → FINISH2(x)

The TRS R consists of the following rules:

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 10 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

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

The TRS R consists of the following rules:

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(12) 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.

(13) Obligation:

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

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

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

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(14) 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].

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

(15) Obligation:

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

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

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

(16) 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(Cb(x))) → BB(moveleft(x))


Used ordering: Polynomial interpretation [POLO]:

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

(17) Obligation:

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

BB(moveleft(Ca(x))) → BB(moveleft(x))

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

(18) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(BB(x1)) = x1   
POL(Ca(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

(19) Obligation:

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) YES

(22) Obligation:

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

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

The TRS R consists of the following rules:

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(23) 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.

(24) Obligation:

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

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

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

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(25) 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].

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

(26) Obligation:

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

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

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

(27) 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:

BA(moveleft(Cb(x))) → BA(moveleft(x))


Used ordering: Polynomial interpretation [POLO]:

POL(BA(x1)) = x1   
POL(Ca(x1)) = 2·x1   
POL(Cb(x1)) = 1 + x1   
POL(moveleft(x1)) = 2·x1   

(28) Obligation:

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

BA(moveleft(Ca(x))) → BA(moveleft(x))

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

(29) 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))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(BA(x1)) = x1   
POL(Ca(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

(30) Obligation:

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) YES

(33) Obligation:

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

FINISH2(Db(x)) → FINISH2(x)
FINISH2(Da(x)) → FINISH2(x)

The TRS R consists of the following rules:

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(34) 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.

(35) Obligation:

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

FINISH2(Db(x)) → FINISH2(x)
FINISH2(Da(x)) → FINISH2(x)

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

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(36) 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].

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

(37) Obligation:

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

FINISH2(Db(x)) → FINISH2(x)
FINISH2(Da(x)) → FINISH2(x)

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

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

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

(39) YES

(40) Obligation:

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

FINISH(Cb(x)) → FINISH(x)
FINISH(Ca(x)) → FINISH(x)

The TRS R consists of the following rules:

a(guess(x)) → guess(Ca(x))
b(guess(x)) → guess(Cb(x))
Ba(moveleft(Ca(x))) → Aa(Ba(moveleft(x)))
Ba(moveleft(Cb(x))) → Ab(Ba(moveleft(x)))
Bb(moveleft(Ca(x))) → Aa(Bb(moveleft(x)))
Bb(moveleft(Cb(x))) → Ab(Bb(moveleft(x)))
Ba(moveleft(cut(x))) → goright(cut(Da(x)))
Bb(moveleft(cut(x))) → goright(cut(Db(x)))
Aa(goright(x)) → goright(Ca(x))
Ab(goright(x)) → goright(Cb(x))
a(wait(goright(x))) → wait(Ba(moveleft(x)))
b(wait(goright(x))) → wait(Bb(moveleft(x)))
finish(Ca(x)) → a(finish(x))
finish(Cb(x)) → b(finish(x))
finish2(Da(x)) → a(finish2(x))
finish2(Db(x)) → b(finish2(x))

The set Q consists of the following terms:

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(41) 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.

(42) Obligation:

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

FINISH(Cb(x)) → FINISH(x)
FINISH(Ca(x)) → FINISH(x)

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

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

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

(43) 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].

a(guess(x0))
b(guess(x0))
Ba(moveleft(Ca(x0)))
Ba(moveleft(Cb(x0)))
Bb(moveleft(Ca(x0)))
Bb(moveleft(Cb(x0)))
Ba(moveleft(cut(x0)))
Bb(moveleft(cut(x0)))
Aa(goright(x0))
Ab(goright(x0))
a(wait(goright(x0)))
b(wait(goright(x0)))
finish(Ca(x0))
finish(Cb(x0))
finish2(Da(x0))
finish2(Db(x0))

(44) Obligation:

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

FINISH(Cb(x)) → FINISH(x)
FINISH(Ca(x)) → FINISH(x)

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

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

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

(46) YES