Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema

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

0 QTRS

↳1 QTRS Reverse (⇔, 0 ms)

↳2 QTRS

↳3 QTRSRRRProof (⇔, 176 ms)

↳4 QTRS

↳5 Overlay + Local Confluence (⇔, 223 ms)

↳6 QTRS

↳7 DependencyPairsProof (⇔, 0 ms)

↳8 QDP

↳9 DependencyGraphProof (⇔, 0 ms)

↳10 QDP

↳11 UsableRulesProof (⇔, 0 ms)

↳12 QDP

↳13 QReductionProof (⇔, 0 ms)

↳14 QDP

↳15 QDPSizeChangeProof (⇔, 0 ms)

↳16 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) 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:

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

Q is empty.

(3) 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:

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

(4) Obligation:

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

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

The set Q consists of the following terms:

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

LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → D(Left(x))
LEFT(Ad(x)) → LEFT(x)
LEFT(Af(x)) → F(Left(x))
LEFT(Af(x)) → LEFT(x)
LEFT(Ag(x)) → G(Left(x))
LEFT(Ag(x)) → LEFT(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
f(Right1(x0))
f(Right2(x0))
f(Right3(x0))
f(Right4(x0))
f(Right5(x0))
f(Right6(x0))
g(Right1(x0))
g(Right2(x0))
g(Right3(x0))
g(Right4(x0))
g(Right5(x0))
g(Right6(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(x0))
Left(Ad(x0))
Left(Af(x0))
Left(Ag(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 1 SCC with 6 less nodes.

(10) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)
LEFT(Af(x)) → LEFT(x)
LEFT(Ag(x)) → LEFT(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(11) 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.

(12) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)
LEFT(Af(x)) → LEFT(x)
LEFT(Ag(x)) → LEFT(x)

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

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

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

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

(14) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)
LEFT(Af(x)) → LEFT(x)
LEFT(Ag(x)) → LEFT(x)

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

(15) 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:

LEFT(Ab(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Aa(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Ac(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Ad(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Af(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Ag(x)) → LEFT(x)
The graph contains the following edges 1 > 1

(0) Obligation:

(1) QTRS Reverse (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) Overlay + Local Confluence (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyPairsProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyGraphProof (EQUIVALENT transformation)

(10) Obligation:

(11) UsableRulesProof (EQUIVALENT transformation)

(12) Obligation:

(13) QReductionProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPSizeChangeProof (EQUIVALENT transformation)

(16) YES