Termination Proof

by AProVE (version ba869e7b28377cd372aedcb96abeb62c4ad6aaa5 rene 20130719 unpublished dirty )

Input

The rewrite relation of the following TRS is considered.

Tlbeta(SRlbeta(x)) → SRlbeta(Tlbeta(x))
Tlbeta(SRlbeta(x)) → SRlbeta(SRlbeta(x))
Tlbeta(SRcp(x)) → SRcp(Tlbeta(x))
Tlbeta(SRcp(x)) → SRlbeta(SRcp(x))
Tlbeta(SRlll(x)) → SRlll(Tlbeta(x))
Tlbeta(SRlll(x)) → SRlbeta(SRlll(x))
Tlbeta(Answer) → Answer
Tlbeta(Answer) → SRlbeta(Answer)
The evaluation strategy is innermost.

Proof

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[Tlbeta(x1)] = 1 · x1 + 1
[SRlbeta(x1)] = 1 · x1 + 1
[SRcp(x1)] = 1 · x1
[SRlll(x1)] = 1 · x1
[Answer] = 0
the rules
Tlbeta(SRlbeta(x)) → SRlbeta(Tlbeta(x))
Tlbeta(SRlbeta(x)) → SRlbeta(SRlbeta(x))
Tlbeta(SRcp(x)) → SRcp(Tlbeta(x))
Tlbeta(SRcp(x)) → SRlbeta(SRcp(x))
Tlbeta(SRlll(x)) → SRlll(Tlbeta(x))
Tlbeta(SRlll(x)) → SRlbeta(SRlll(x))
Tlbeta(Answer) → SRlbeta(Answer)
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[Tlbeta(x1)] = 1 · x1 + 1
[SRlbeta(x1)] = 1 · x1
[SRcp(x1)] = 1 · x1
[SRlll(x1)] = 1 · x1
[Answer] = 0
the rules
Tlbeta(SRlbeta(x)) → SRlbeta(Tlbeta(x))
Tlbeta(SRcp(x)) → SRcp(Tlbeta(x))
Tlbeta(SRlll(x)) → SRlll(Tlbeta(x))
remain.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
Tlbeta#(SRlbeta(x)) → Tlbeta#(x)
Tlbeta#(SRcp(x)) → Tlbeta#(x)
Tlbeta#(SRlll(x)) → Tlbeta#(x)

1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

As there is no critical graph in the transitive closure, there are no infinite chains.