Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z026)

The rewrite relation of the following TRS is considered.

a(b(b(x1)))	→	b(a(a(x1)))	(1)
a(a(b(x1)))	→	b(b(a(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.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(b(x1))) → a^#(x1) (5)

a^#(b(b(x1))) → a^#(x1) (4)

a^#(b(b(x1))) → a^#(a(x1)) (3)

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

[a(x₁)] = x₁ + 11798

[b(x₁)] = x₁ + 11798

[a^#(x₁)] = x₁ + 0

together with the usable rules

a(b(b(x1))) → b(a(a(x1))) (1)

a(a(b(x1))) → b(b(a(x1))) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pairs

a^#(a(b(x1))) → a^#(x1) (5)

a^#(b(b(x1))) → a^#(x1) (4)

a^#(b(b(x1))) → a^#(a(x1)) (3)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.