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 NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

b^#(r(x)) → b^#(x) (6)

b^#(w(x)) → b^#(x) (5)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[w^#(x₁)] = 0

[b(x₁)] = 0

[r(x₁)] = x₁ + 1

[w(x₁)] = x₁ + 1

[b^#(x₁)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

b^#(r(x)) → b^#(x) (6)

b^#(w(x)) → b^#(x) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.
The 2^nd component contains the pair
w^#(r(x)) → w^#(x) (7)

1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[w^#(x₁)] = x₁ + 0

[b(x₁)] = 0

[r(x₁)] = x₁ + 1

[w(x₁)] = 1

[b^#(x₁)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
w^#(r(x)) → w^#(x) (7)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.