YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z102-split.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 61 ms)
↳4 QTRS
↳5 Overlay + Local Confluence (⇔, 10 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(c(c(x)))) → Wait(Right3(x))
Begin(c(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(b(b(b(End(x)))))
Right2(b(End(x))) → Left(c(c(c(End(x)))))
Right3(c(End(x))) → Left(a(b(End(x))))
Right4(c(c(End(x)))) → Left(a(b(End(x))))
Right5(c(c(c(End(x))))) → Left(a(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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(x)) → b(b(b(x)))
b(b(x)) → c(c(c(x)))
c(c(c(c(x)))) → a(b(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(c(c(Begin(x)))) → Right3(Wait(x))
c(c(Begin(x))) → Right4(Wait(x))
c(Begin(x)) → Right5(Wait(x))
End(a(Right1(x))) → End(b(b(b(Left(x)))))
End(b(Right2(x))) → End(c(c(c(Left(x)))))
End(c(Right3(x))) → End(b(a(Left(x))))
End(c(c(Right4(x)))) → End(b(a(Left(x))))
End(c(c(c(Right5(x))))) → End(b(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))
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))
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))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(a(x)) → b(b(b(x)))
b(b(x)) → c(c(c(x)))
c(c(c(c(x)))) → b(a(x))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 78 + x1   
POL(Ab(x1)) = 51 + x1   
POL(Ac(x1)) = 33 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 76 + x1   
POL(Right2(x1)) = 49 + x1   
POL(Right3(x1)) = 97 + x1   
POL(Right4(x1)) = 64 + x1   
POL(Right5(x1)) = 31 + x1   
POL(Wait(x1)) = 1 + x1   
POL(a(x1)) = 78 + x1   
POL(b(x1)) = 51 + x1   
POL(c(x1)) = 33 + 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(c(c(Begin(x)))) → Right3(Wait(x))
c(c(Begin(x))) → Right4(Wait(x))
c(Begin(x)) → Right5(Wait(x))
End(a(Right1(x))) → End(b(b(b(Left(x)))))
End(b(Right2(x))) → End(c(c(c(Left(x)))))
End(c(Right3(x))) → End(b(a(Left(x))))
End(c(c(Right4(x)))) → End(b(a(Left(x))))
End(c(c(c(Right5(x))))) → End(b(a(Left(x))))
Left(Wait(x)) → Begin(x)
a(a(x)) → b(b(b(x)))
b(b(x)) → c(c(c(x)))
c(c(c(c(x)))) → b(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))
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))
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))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ac(x)) → c(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))
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))
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))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ac(x)) → c(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))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(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)

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))
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))
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))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ac(x)) → c(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))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(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 3 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)

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))
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))
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))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ac(x)) → c(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))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(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)

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))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(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))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(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)

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

(16) YES