YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z104-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 (⇔, 88 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 17 ms)
↳6 QTRS
↳7 QTRSRRRProof (⇔, 15 ms)
↳8 QTRS
↳9 QTRSRRRProof (⇔, 17 ms)
↳10 QTRS
↳11 QTRSRRRProof (⇔, 0 ms)
↳12 QTRS
↳13 Overlay + Local Confluence (⇔, 0 ms)
↳14 QTRS
↳15 DependencyPairsProof (⇔, 13 ms)
↳16 QDP
↳17 DependencyGraphProof (⇔, 1 ms)
↳18 QDP
↳19 UsableRulesProof (⇔, 0 ms)
↳20 QDP
↳21 QReductionProof (⇔, 0 ms)
↳22 QDP
↳23 QDPSizeChangeProof (⇔, 0 ms)
↳24 YES

(0) Obligation:

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

Begin(c(c(a(x)))) → Wait(Right1(x))
Begin(c(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Begin(d(x)) → Wait(Right5(x))
Begin(c(c(c(x)))) → Wait(Right6(x))
Begin(c(c(x))) → Wait(Right7(x))
Begin(c(x)) → Wait(Right8(x))
Right1(c(End(x))) → Left(d(d(End(x))))
Right2(c(c(End(x)))) → Left(d(d(End(x))))
Right3(c(c(c(End(x))))) → Left(d(d(End(x))))
Right4(d(End(x))) → Left(c(c(End(x))))
Right5(b(End(x))) → Left(c(c(End(x))))
Right6(a(End(x))) → Left(d(d(End(x))))
Right7(a(c(End(x)))) → Left(d(d(End(x))))
Right8(a(c(c(End(x))))) → Left(d(d(End(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))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(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))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(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))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(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))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Ad(Left(x)) → Left(d(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
c(c(c(a(x)))) → d(d(x))
d(b(x)) → c(c(x))
c(x) → a(a(a(a(x))))
d(x) → b(b(b(b(x))))
b(d(x)) → c(c(x))
a(c(c(c(x)))) → d(d(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(c(c(Begin(x)))) → Right1(Wait(x))
a(c(Begin(x))) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
c(c(c(Begin(x)))) → Right6(Wait(x))
c(c(Begin(x))) → Right7(Wait(x))
c(Begin(x)) → Right8(Wait(x))
End(c(Right1(x))) → End(d(d(Left(x))))
End(c(c(Right2(x)))) → End(d(d(Left(x))))
End(c(c(c(Right3(x))))) → End(d(d(Left(x))))
End(d(Right4(x))) → End(c(c(Left(x))))
End(b(Right5(x))) → End(c(c(Left(x))))
End(a(Right6(x))) → End(d(d(Left(x))))
End(c(a(Right7(x)))) → End(d(d(Left(x))))
End(c(c(a(Right8(x))))) → End(d(d(Left(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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
c(x) → a(a(a(a(x))))
d(x) → b(b(b(b(x))))
d(b(x)) → c(c(x))
c(c(c(a(x)))) → d(d(x))

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 3 + x1   
POL(Ab(x1)) = 5 + x1   
POL(Ac(x1)) = 13 + x1   
POL(Ad(x1)) = 21 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 29 + x1   
POL(Right2(x1)) = 16 + x1   
POL(Right3(x1)) = 3 + x1   
POL(Right4(x1)) = 5 + x1   
POL(Right5(x1)) = 21 + x1   
POL(Right6(x1)) = 39 + x1   
POL(Right7(x1)) = 26 + x1   
POL(Right8(x1)) = 13 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 3 + x1   
POL(b(x1)) = 5 + x1   
POL(c(x1)) = 13 + x1   
POL(d(x1)) = 21 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(4) Obligation:

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

a(c(c(Begin(x)))) → Right1(Wait(x))
a(c(Begin(x))) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
c(c(c(Begin(x)))) → Right6(Wait(x))
c(c(Begin(x))) → Right7(Wait(x))
c(Begin(x)) → Right8(Wait(x))
End(c(Right1(x))) → End(d(d(Left(x))))
End(c(c(Right2(x)))) → End(d(d(Left(x))))
End(c(c(c(Right3(x))))) → End(d(d(Left(x))))
End(d(Right4(x))) → End(c(c(Left(x))))
End(b(Right5(x))) → End(c(c(Left(x))))
End(a(Right6(x))) → End(d(d(Left(x))))
End(c(a(Right7(x)))) → End(d(d(Left(x))))
End(c(c(a(Right8(x))))) → End(d(d(Left(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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(c(c(c(x)))) → d(d(x))
b(d(x)) → c(c(x))
d(b(x)) → c(c(x))
c(c(c(a(x)))) → d(d(x))

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = 1 + x1   
POL(Ac(x1)) = x1   
POL(Ad(x1)) = x1   
POL(Begin(x1)) = 1 + x1   
POL(End(x1)) = x1   
POL(Left(x1)) = 1 + x1   
POL(Right1(x1)) = 1 + x1   
POL(Right2(x1)) = 1 + x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = 1 + x1   
POL(Right5(x1)) = x1   
POL(Right6(x1)) = 1 + x1   
POL(Right7(x1)) = 1 + x1   
POL(Right8(x1)) = 1 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

b(Begin(x)) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
b(d(x)) → c(c(x))
d(b(x)) → c(c(x))


(6) Obligation:

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

a(c(c(Begin(x)))) → Right1(Wait(x))
a(c(Begin(x))) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
c(c(c(Begin(x)))) → Right6(Wait(x))
c(c(Begin(x))) → Right7(Wait(x))
c(Begin(x)) → Right8(Wait(x))
End(c(Right1(x))) → End(d(d(Left(x))))
End(c(c(Right2(x)))) → End(d(d(Left(x))))
End(c(c(c(Right3(x))))) → End(d(d(Left(x))))
End(d(Right4(x))) → End(c(c(Left(x))))
End(b(Right5(x))) → End(c(c(Left(x))))
End(a(Right6(x))) → End(d(d(Left(x))))
End(c(a(Right7(x)))) → End(d(d(Left(x))))
End(c(c(a(Right8(x))))) → End(d(d(Left(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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(c(c(c(x)))) → d(d(x))
c(c(c(a(x)))) → d(d(x))

Q is empty.

(7) 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(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = x1   
POL(Right4(x1)) = 1 + x1   
POL(Right5(x1)) = 1 + x1   
POL(Right6(x1)) = x1   
POL(Right7(x1)) = x1   
POL(Right8(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

End(d(Right4(x))) → End(c(c(Left(x))))
End(b(Right5(x))) → End(c(c(Left(x))))


(8) Obligation:

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

a(c(c(Begin(x)))) → Right1(Wait(x))
a(c(Begin(x))) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
c(c(c(Begin(x)))) → Right6(Wait(x))
c(c(Begin(x))) → Right7(Wait(x))
c(Begin(x)) → Right8(Wait(x))
End(c(Right1(x))) → End(d(d(Left(x))))
End(c(c(Right2(x)))) → End(d(d(Left(x))))
End(c(c(c(Right3(x))))) → End(d(d(Left(x))))
End(a(Right6(x))) → End(d(d(Left(x))))
End(c(a(Right7(x)))) → End(d(d(Left(x))))
End(c(c(a(Right8(x))))) → End(d(d(Left(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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(c(c(c(x)))) → d(d(x))
c(c(c(a(x)))) → d(d(x))

Q is empty.

(9) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 3 + x1   
POL(Ab(x1)) = x1   
POL(Ac(x1)) = x1   
POL(Ad(x1)) = x1   
POL(Begin(x1)) = 1 + x1   
POL(End(x1)) = x1   
POL(Left(x1)) = 2 + x1   
POL(Right1(x1)) = 3 + x1   
POL(Right2(x1)) = 4 + x1   
POL(Right3(x1)) = 2 + x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = x1   
POL(Right6(x1)) = 1 + x1   
POL(Right7(x1)) = x1   
POL(Right8(x1)) = 1 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 3 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a(c(c(Begin(x)))) → Right1(Wait(x))
a(Begin(x)) → Right3(Wait(x))
c(c(Begin(x))) → Right7(Wait(x))
End(c(Right1(x))) → End(d(d(Left(x))))
End(c(c(Right2(x)))) → End(d(d(Left(x))))
End(a(Right6(x))) → End(d(d(Left(x))))
End(c(a(Right7(x)))) → End(d(d(Left(x))))
End(c(c(a(Right8(x))))) → End(d(d(Left(x))))
Left(Wait(x)) → Begin(x)
a(c(c(c(x)))) → d(d(x))
c(c(c(a(x)))) → d(d(x))


(10) Obligation:

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

a(c(Begin(x))) → Right2(Wait(x))
c(c(c(Begin(x)))) → Right6(Wait(x))
c(Begin(x)) → Right8(Wait(x))
End(c(c(c(Right3(x))))) → End(d(d(Left(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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))

Q is empty.

(11) 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(Begin(x1)) = 1 + x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = x1   
POL(Right6(x1)) = x1   
POL(Right7(x1)) = x1   
POL(Right8(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a(c(Begin(x))) → Right2(Wait(x))
c(c(c(Begin(x)))) → Right6(Wait(x))
c(Begin(x)) → Right8(Wait(x))
End(c(c(c(Right3(x))))) → End(d(d(Left(x))))


(12) Obligation:

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

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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))

Q is empty.

(13) Overlay + Local Confluence (EQUIVALENT transformation)

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

(14) Obligation:

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

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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))

The set Q consists of the following terms:

c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
c(Right7(x0))
c(Right8(x0))
a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
a(Right7(x0))
a(Right8(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
d(Right7(x0))
d(Right8(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
b(Right7(x0))
b(Right8(x0))
Left(Ac(x0))
Left(Aa(x0))
Left(Ad(x0))
Left(Ab(x0))

(15) DependencyPairsProof (EQUIVALENT transformation)

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

(16) Obligation:

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

LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ad(x)) → D(Left(x))
LEFT(Ad(x)) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)

The TRS R consists of the following rules:

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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))

The set Q consists of the following terms:

c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
c(Right7(x0))
c(Right8(x0))
a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
a(Right7(x0))
a(Right8(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
d(Right7(x0))
d(Right8(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
b(Right7(x0))
b(Right8(x0))
Left(Ac(x0))
Left(Aa(x0))
Left(Ad(x0))
Left(Ab(x0))

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(18) Obligation:

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

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

The TRS R consists of the following rules:

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))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(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))
a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(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))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(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))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
Left(Ac(x)) → c(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ab(x)) → b(Left(x))

The set Q consists of the following terms:

c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
c(Right7(x0))
c(Right8(x0))
a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
a(Right7(x0))
a(Right8(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
d(Right7(x0))
d(Right8(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
b(Right7(x0))
b(Right8(x0))
Left(Ac(x0))
Left(Aa(x0))
Left(Ad(x0))
Left(Ab(x0))

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

(19) 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.

(20) Obligation:

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

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

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

c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
c(Right7(x0))
c(Right8(x0))
a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
a(Right7(x0))
a(Right8(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
d(Right7(x0))
d(Right8(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
b(Right7(x0))
b(Right8(x0))
Left(Ac(x0))
Left(Aa(x0))
Left(Ad(x0))
Left(Ab(x0))

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

(21) 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].

c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
c(Right7(x0))
c(Right8(x0))
a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
a(Right7(x0))
a(Right8(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
d(Right7(x0))
d(Right8(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
b(Right7(x0))
b(Right8(x0))
Left(Ac(x0))
Left(Aa(x0))
Left(Ad(x0))
Left(Ab(x0))

(22) Obligation:

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

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

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

(23) 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(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(Ab(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

(24) YES