Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z006)

The rewrite relation of the following TRS is considered.

a(b(x1))	→	b(a(x1))	(1)
b(a(x1))	→	a(c(b(x1)))	(2)

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^#(a(x1))	→	b^#(x1)	(3)
a^#(b(x1))	→	b^#(a(x1))	(4)
b^#(a(x1))	→	a^#(c(b(x1)))	(5)
a^#(b(x1))	→	a^#(x1)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
a^#(b(x1)) → a^#(x1) (6)

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

[a(x₁)] = 0

[b(x₁)] = x₁ + 1

[c(x₁)] = 0

[a^#(x₁)] = x₁ + 0

[b^#(x₁)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
a^#(b(x1)) → a^#(x1) (6)
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
b^#(a(x1)) → b^#(x1) (3)

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

[a(x₁)] = x₁ + 1

[b(x₁)] = 1

[c(x₁)] = 0

[a^#(x₁)] = 0

[b^#(x₁)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
b^#(a(x1)) → b^#(x1) (3)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.