YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z118-split.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 64 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 6 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 10 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 10 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 QDP
        ↳17 UsableRulesProof (⇔, 1 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 1 ms)
        ↳23 QDP
        ↳24 QDPSizeChangeProof (⇔, 0 ms)
        ↳25 YES
    ↳26 QDP
        ↳27 UsableRulesProof (⇔, 2 ms)
        ↳28 QDP
        ↳29 QDPSizeChangeProof (⇔, 0 ms)
        ↳30 YES
    ↳31 QDP
        ↳32 UsableRulesProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 QDPSizeChangeProof (⇔, 0 ms)
        ↳35 YES
    ↳36 QDP
        ↳37 UsableRulesProof (⇔, 1 ms)
        ↳38 QDP
        ↳39 QDPSizeChangeProof (⇔, 0 ms)
        ↳40 YES
    ↳41 QDP
        ↳42 UsableRulesProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 QDPSizeChangeProof (⇔, 0 ms)
        ↳45 YES
    ↳46 QDP
        ↳47 UsableRulesProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 QDPOrderProof (⇔, 22 ms)
        ↳50 QDP
        ↳51 DependencyGraphProof (⇔, 0 ms)
        ↳52 TRUE

(0) Obligation:

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

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(b(x)) → Wait(Right3(x))
Begin(d(x)) → Wait(Right4(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right1(b(End(x))) → Left(c(d(c(End(x)))))
Right2(b(b(End(x)))) → Left(c(d(c(End(x)))))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right4(c(End(x))) → Left(g(g(End(x))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(b(x))) → c(d(c(x)))
b(b(x)) → a(g(g(x)))
c(d(x)) → g(g(x))
g(g(g(x))) → b(b(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 12 + x1   
POL(Ab(x1)) = 18 + x1   
POL(Ac(x1)) = 26 + x1   
POL(Ad(x1)) = x1   
POL(Ag(x1)) = 12 + x1   
POL(Begin(x1)) = 1 + x1   
POL(End(x1)) = x1   
POL(Left(x1)) = 1 + x1   
POL(Right1(x1)) = 37 + x1   
POL(Right2(x1)) = 18 + x1   
POL(Right3(x1)) = 19 + x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = 25 + x1   
POL(Right6(x1)) = 13 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 12 + x1   
POL(b(x1)) = 18 + x1   
POL(c(x1)) = 26 + x1   
POL(d(x1)) = x1   
POL(g(x1)) = 12 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Begin(b(x)) → Wait(Right2(x))
Begin(d(x)) → Wait(Right4(x))
Right1(b(End(x))) → Left(c(d(c(End(x)))))
Right2(b(b(End(x)))) → Left(c(d(c(End(x)))))
Right4(c(End(x))) → Left(g(g(End(x))))
b(b(b(x))) → c(d(c(x)))
c(d(x)) → g(g(x))


(2) Obligation:

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

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

Q is empty.

(3) 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(Ag(x1)) = x1   
POL(Begin(x1)) = 1 + x1   
POL(End(x1)) = x1   
POL(Left(x1)) = 1 + x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = 1 + x1   
POL(Right6(x1)) = 1 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(g(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Begin(b(b(x))) → Wait(Right1(x))


(4) Obligation:

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

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

BEGIN(b(x)) → WAIT(Right3(x))
BEGIN(b(x)) → RIGHT3(x)
BEGIN(g(g(x))) → WAIT(Right5(x))
BEGIN(g(g(x))) → RIGHT5(x)
BEGIN(g(x)) → WAIT(Right6(x))
BEGIN(g(x)) → RIGHT6(x)
RIGHT3(b(End(x))) → A(g(g(End(x))))
RIGHT3(b(End(x))) → G(g(End(x)))
RIGHT3(b(End(x))) → G(End(x))
RIGHT5(g(End(x))) → B(b(End(x)))
RIGHT5(g(End(x))) → B(End(x))
RIGHT6(g(g(End(x)))) → B(b(End(x)))
RIGHT6(g(g(End(x)))) → B(End(x))
RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
RIGHT4(a(x)) → AA(Right4(x))
RIGHT4(a(x)) → RIGHT4(x)
RIGHT5(a(x)) → AA(Right5(x))
RIGHT5(a(x)) → RIGHT5(x)
RIGHT6(a(x)) → AA(Right6(x))
RIGHT6(a(x)) → RIGHT6(x)
RIGHT1(g(x)) → AG(Right1(x))
RIGHT1(g(x)) → RIGHT1(x)
RIGHT2(g(x)) → AG(Right2(x))
RIGHT2(g(x)) → RIGHT2(x)
RIGHT3(g(x)) → AG(Right3(x))
RIGHT3(g(x)) → RIGHT3(x)
RIGHT4(g(x)) → AG(Right4(x))
RIGHT4(g(x)) → RIGHT4(x)
RIGHT5(g(x)) → AG(Right5(x))
RIGHT5(g(x)) → RIGHT5(x)
RIGHT6(g(x)) → AG(Right6(x))
RIGHT6(g(x)) → RIGHT6(x)
RIGHT1(d(x)) → AD(Right1(x))
RIGHT1(d(x)) → RIGHT1(x)
RIGHT2(d(x)) → AD(Right2(x))
RIGHT2(d(x)) → RIGHT2(x)
RIGHT3(d(x)) → AD(Right3(x))
RIGHT3(d(x)) → RIGHT3(x)
RIGHT4(d(x)) → AD(Right4(x))
RIGHT4(d(x)) → RIGHT4(x)
RIGHT5(d(x)) → AD(Right5(x))
RIGHT5(d(x)) → RIGHT5(x)
RIGHT6(d(x)) → AD(Right6(x))
RIGHT6(d(x)) → RIGHT6(x)
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT4(b(x)) → AB(Right4(x))
RIGHT4(b(x)) → RIGHT4(x)
RIGHT5(b(x)) → AB(Right5(x))
RIGHT5(b(x)) → RIGHT5(x)
RIGHT6(b(x)) → AB(Right6(x))
RIGHT6(b(x)) → RIGHT6(x)
RIGHT1(c(x)) → AC(Right1(x))
RIGHT1(c(x)) → RIGHT1(x)
RIGHT2(c(x)) → AC(Right2(x))
RIGHT2(c(x)) → RIGHT2(x)
RIGHT3(c(x)) → AC(Right3(x))
RIGHT3(c(x)) → RIGHT3(x)
RIGHT4(c(x)) → AC(Right4(x))
RIGHT4(c(x)) → RIGHT4(x)
RIGHT5(c(x)) → AC(Right5(x))
RIGHT5(c(x)) → RIGHT5(x)
RIGHT6(c(x)) → AC(Right6(x))
RIGHT6(c(x)) → RIGHT6(x)
AA(Left(x)) → A(x)
AG(Left(x)) → G(x)
AB(Left(x)) → B(x)
WAIT(Left(x)) → BEGIN(x)
A(x) → G(d(x))
B(b(x)) → A(g(g(x)))
B(b(x)) → G(g(x))
B(b(x)) → G(x)
G(g(g(x))) → B(b(x))
G(g(g(x))) → B(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 45 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

B(b(x)) → G(g(x))
G(g(g(x))) → B(b(x))
B(b(x)) → G(x)
G(g(g(x))) → B(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(10) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(11) Obligation:

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

B(b(x)) → G(g(x))
G(g(g(x))) → B(b(x))
B(b(x)) → G(x)
G(g(g(x))) → B(x)

The TRS R consists of the following rules:

b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))
a(x) → g(d(x))

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

(12) 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:

B(b(x)) → G(g(x))
G(g(g(x))) → B(b(x))
B(b(x)) → G(x)
G(g(g(x))) → B(x)


Used ordering: Polynomial interpretation [POLO]:

POL(B(x1)) = 2·x1   
POL(G(x1)) = 2·x1   
POL(a(x1)) = 2 + x1   
POL(b(x1)) = 3 + x1   
POL(d(x1)) = x1   
POL(g(x1)) = 2 + x1   

(13) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))
a(x) → g(d(x))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) YES

(16) Obligation:

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

RIGHT4(g(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)
RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(c(x)) → RIGHT4(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(17) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(18) Obligation:

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

RIGHT4(g(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)
RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(c(x)) → RIGHT4(x)

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

(19) 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:

  • RIGHT4(g(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(a(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(d(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(b(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(c(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

RIGHT2(g(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(c(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(22) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

RIGHT2(g(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(c(x)) → RIGHT2(x)

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

(24) 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:

  • RIGHT2(g(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(a(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(d(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(b(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(c(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

(25) YES

(26) Obligation:

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

RIGHT1(g(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(c(x)) → RIGHT1(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(27) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(28) Obligation:

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

RIGHT1(g(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(c(x)) → RIGHT1(x)

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

(29) 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:

  • RIGHT1(g(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(a(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(d(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(b(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(c(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

(30) YES

(31) Obligation:

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

RIGHT6(g(x)) → RIGHT6(x)
RIGHT6(a(x)) → RIGHT6(x)
RIGHT6(d(x)) → RIGHT6(x)
RIGHT6(b(x)) → RIGHT6(x)
RIGHT6(c(x)) → RIGHT6(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(32) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(33) Obligation:

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

RIGHT6(g(x)) → RIGHT6(x)
RIGHT6(a(x)) → RIGHT6(x)
RIGHT6(d(x)) → RIGHT6(x)
RIGHT6(b(x)) → RIGHT6(x)
RIGHT6(c(x)) → RIGHT6(x)

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

(34) 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:

  • RIGHT6(g(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(a(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(d(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(b(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(c(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

(35) YES

(36) Obligation:

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

RIGHT5(g(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)
RIGHT5(d(x)) → RIGHT5(x)
RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(c(x)) → RIGHT5(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(37) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(38) Obligation:

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

RIGHT5(g(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)
RIGHT5(d(x)) → RIGHT5(x)
RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(c(x)) → RIGHT5(x)

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

(39) 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:

  • RIGHT5(g(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(a(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(d(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(b(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(c(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

(40) YES

(41) Obligation:

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

RIGHT3(g(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(c(x)) → RIGHT3(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(42) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(43) Obligation:

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

RIGHT3(g(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(c(x)) → RIGHT3(x)

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

(44) 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:

  • RIGHT3(g(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(a(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(d(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(b(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(c(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

(45) YES

(46) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(x)) → WAIT(Right3(x))
BEGIN(g(g(x))) → WAIT(Right5(x))
BEGIN(g(x)) → WAIT(Right6(x))

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right3(x))
Begin(g(g(x))) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right5(g(End(x))) → Left(b(b(End(x))))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right1(g(x)) → Ag(Right1(x))
Right2(g(x)) → Ag(Right2(x))
Right3(g(x)) → Ag(Right3(x))
Right4(g(x)) → Ag(Right4(x))
Right5(g(x)) → Ag(Right5(x))
Right6(g(x)) → Ag(Right6(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Aa(Left(x)) → Left(a(x))
Ag(Left(x)) → Left(g(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(x) → g(d(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))

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

(47) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(48) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(x)) → WAIT(Right3(x))
BEGIN(g(g(x))) → WAIT(Right5(x))
BEGIN(g(x)) → WAIT(Right6(x))

The TRS R consists of the following rules:

Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right6(a(x)) → Aa(Right6(x))
Right6(g(x)) → Ag(Right6(x))
Right6(d(x)) → Ad(Right6(x))
Right6(b(x)) → Ab(Right6(x))
Right6(c(x)) → Ac(Right6(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))
a(x) → g(d(x))
Ad(Left(x)) → Left(d(x))
Ag(Left(x)) → Left(g(x))
Aa(Left(x)) → Left(a(x))
Right5(g(End(x))) → Left(b(b(End(x))))
Right5(a(x)) → Aa(Right5(x))
Right5(g(x)) → Ag(Right5(x))
Right5(d(x)) → Ad(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right3(a(x)) → Aa(Right3(x))
Right3(g(x)) → Ag(Right3(x))
Right3(d(x)) → Ad(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))

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

(49) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


BEGIN(b(x)) → WAIT(Right3(x))
BEGIN(g(g(x))) → WAIT(Right5(x))
BEGIN(g(x)) → WAIT(Right6(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = 1   
POL(Ab(x1)) = 1 + x1   
POL(Ac(x1)) = 0   
POL(Ad(x1)) = 0   
POL(Ag(x1)) = 1 + x1   
POL(BEGIN(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right3(x1)) = x1   
POL(Right5(x1)) = 1 + x1   
POL(Right6(x1)) = x1   
POL(WAIT(x1)) = x1   
POL(a(x1)) = 1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 0   
POL(d(x1)) = 0   
POL(g(x1)) = 1 + x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right3(a(x)) → Aa(Right3(x))
Right3(g(x)) → Ag(Right3(x))
Right3(d(x)) → Ad(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right5(g(End(x))) → Left(b(b(End(x))))
Right5(a(x)) → Aa(Right5(x))
Right5(g(x)) → Ag(Right5(x))
Right5(d(x)) → Ad(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right6(a(x)) → Aa(Right6(x))
Right6(g(x)) → Ag(Right6(x))
Right6(d(x)) → Ad(Right6(x))
Right6(b(x)) → Ab(Right6(x))
Right6(c(x)) → Ac(Right6(x))
Ag(Left(x)) → Left(g(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))
a(x) → g(d(x))

(50) Obligation:

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

WAIT(Left(x)) → BEGIN(x)

The TRS R consists of the following rules:

Right6(g(g(End(x)))) → Left(b(b(End(x))))
Right6(a(x)) → Aa(Right6(x))
Right6(g(x)) → Ag(Right6(x))
Right6(d(x)) → Ad(Right6(x))
Right6(b(x)) → Ab(Right6(x))
Right6(c(x)) → Ac(Right6(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(b(x)) → a(g(g(x)))
g(g(g(x))) → b(b(x))
a(x) → g(d(x))
Ad(Left(x)) → Left(d(x))
Ag(Left(x)) → Left(g(x))
Aa(Left(x)) → Left(a(x))
Right5(g(End(x))) → Left(b(b(End(x))))
Right5(a(x)) → Aa(Right5(x))
Right5(g(x)) → Ag(Right5(x))
Right5(d(x)) → Ad(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right3(b(End(x))) → Left(a(g(g(End(x)))))
Right3(a(x)) → Aa(Right3(x))
Right3(g(x)) → Ag(Right3(x))
Right3(d(x)) → Ad(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))

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

(51) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(52) TRUE