Certification Problem

Input (TPDB TRS_Standard/Der95/07)

The rewrite relation of the following TRS is considered.

w(r(x))	→	r(w(x))	(1)
b(r(x))	→	r(b(x))	(2)
b(w(x))	→	w(b(x))	(3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

w^#(r(x))	→	w^#(x)	(4)
b^#(r(x))	→	b^#(x)	(5)
b^#(w(x))	→	w^#(b(x))	(6)
b^#(w(x))	→	b^#(x)	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

b^#(w(x)) → b^#(x) (7)

b^#(r(x)) → b^#(x) (5)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[w(x₁)] = 1 · x₁

[r(x₁)] = 1 · x₁

[b^#(x₁)] = 1 · x₁

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
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.

b^#(w(x)) → b^#(x) (7)

1 > 1

b^#(r(x)) → b^#(x) (5)

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.
The 2^nd component contains the pair
w^#(r(x)) → w^#(x) (4)

1.1.2 Switch to Innermost Termination
The TRS does not have overlaps with the pairs and is locally confluent:
20
Hence, it suffices to show innermost termination in the following.
1.1.2.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
1.1.2.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
There are no lhss.
1.1.2.1.1.1 Size-Change Termination
Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

w^#(r(x)) → w^#(x) (4)

1 > 1

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