YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z112-split.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 110 ms)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔, 37 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 48 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QReductionProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPSizeChangeProof (⇔, 0 ms)
        ↳15 YES
    ↳16 QDP
        ↳17 UsableRulesProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QReductionProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QReductionProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QDPSizeChangeProof (⇔, 0 ms)
        ↳29 YES
    ↳30 QDP
        ↳31 UsableRulesProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QReductionProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QReductionProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QDPSizeChangeProof (⇔, 0 ms)
        ↳43 YES
    ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QReductionProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 QDPSizeChangeProof (⇔, 0 ms)
        ↳50 YES

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 465 + x1   
POL(Ab(x1)) = 486 + x1   
POL(Ac(x1)) = 441 + x1   
POL(Ad(x1)) = 528 + x1   
POL(Af(x1)) = 351 + x1   
POL(Ag(x1)) = 234 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 463 + x1   
POL(Right2(x1)) = 484 + x1   
POL(Right3(x1)) = 439 + x1   
POL(Right4(x1)) = 526 + x1   
POL(Right5(x1)) = 349 + x1   
POL(Right6(x1)) = 232 + x1   
POL(Wait(x1)) = 1 + x1   
POL(a(x1)) = 465 + x1   
POL(b(x1)) = 486 + x1   
POL(c(x1)) = 441 + x1   
POL(d(x1)) = 528 + x1   
POL(f(x1)) = 351 + x1   
POL(g(x1)) = 234 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Begin(a(x)) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Begin(d(x)) → Wait(Right4(x))
Begin(f(x)) → Wait(Right5(x))
Begin(g(x)) → Wait(Right6(x))
Right1(a(End(x))) → Left(b(c(End(x))))
Right2(b(End(x))) → Left(c(d(End(x))))
Right3(c(End(x))) → Left(d(f(End(x))))
Right4(d(End(x))) → Left(f(f(f(End(x)))))
Right5(f(End(x))) → Left(g(a(End(x))))
Right6(g(End(x))) → Left(a(End(x)))
Wait(Left(x)) → Begin(x)
a(a(x)) → b(c(x))
b(b(x)) → c(d(x))
b(x) → a(x)
c(c(x)) → d(f(x))
d(d(x)) → f(f(f(x)))
d(x) → b(x)
f(f(x)) → g(a(x))
g(g(x)) → a(x)


(2) Obligation:

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

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

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(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:

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(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)
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(f(x)) → AF(Right1(x))
RIGHT1(f(x)) → RIGHT1(x)
RIGHT2(f(x)) → AF(Right2(x))
RIGHT2(f(x)) → RIGHT2(x)
RIGHT3(f(x)) → AF(Right3(x))
RIGHT3(f(x)) → RIGHT3(x)
RIGHT4(f(x)) → AF(Right4(x))
RIGHT4(f(x)) → RIGHT4(x)
RIGHT5(f(x)) → AF(Right5(x))
RIGHT5(f(x)) → RIGHT5(x)
RIGHT6(f(x)) → AF(Right6(x))
RIGHT6(f(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)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 36 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(10) 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.

(11) Obligation:

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

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

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

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(12) 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].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(13) Obligation:

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

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

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

(14) 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(b(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(c(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(f(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

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

(15) YES

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(17) 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.

(18) Obligation:

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

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

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

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(19) 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].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(20) Obligation:

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

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

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

(21) 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(b(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(c(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(f(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

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

(22) YES

(23) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(24) 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.

(25) Obligation:

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

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

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

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(26) 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].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(27) Obligation:

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

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

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

(28) 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(b(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(c(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(f(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

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

(29) YES

(30) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(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:

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

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

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(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].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(34) Obligation:

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

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

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

(35) 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(b(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(c(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(f(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

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

(36) YES

(37) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(38) 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.

(39) Obligation:

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

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

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

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(40) 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].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(41) Obligation:

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

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

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

(42) 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(b(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(c(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(f(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

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

(43) YES

(44) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(45) 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.

(46) Obligation:

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

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

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

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

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

(47) 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].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right1(f(x0))
Right2(f(x0))
Right3(f(x0))
Right4(f(x0))
Right5(f(x0))
Right6(f(x0))
Right1(g(x0))
Right2(g(x0))
Right3(g(x0))
Right4(g(x0))
Right5(g(x0))
Right6(g(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))
Af(Left(x0))
Ag(Left(x0))

(48) Obligation:

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

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

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

(49) 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(b(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(c(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(f(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

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

(50) YES