Certification Problem

Input (TPDB TRS_Standard/SK90/2.46)

The rewrite relation of the following TRS is considered.

a(a(x))	→	b(b(x))	(1)
b(b(a(x)))	→	a(b(b(x)))	(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^#(b(a(x)))	→	b^#(b(x))	(3)
a^#(a(x))	→	b^#(x)	(4)
b^#(b(a(x)))	→	b^#(x)	(5)
b^#(b(a(x)))	→	a^#(b(b(x)))	(6)
a^#(a(x))	→	b^#(b(x))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(x)) → b^#(b(x)) (7)

b^#(b(a(x))) → a^#(b(b(x))) (6)

b^#(b(a(x))) → b^#(x) (5)

a^#(a(x)) → b^#(x) (4)

b^#(b(a(x))) → b^#(b(x)) (3)

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

[a(x₁)] = x₁ + 7721

[b(x₁)] = x₁ + 7720

[a^#(x₁)] = x₁ + 0

[b^#(x₁)] = x₁ + 0

together with the usable rules

a(a(x)) → b(b(x)) (1)

b(b(a(x))) → a(b(b(x))) (2)

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

a^#(a(x)) → b^#(b(x)) (7)

b^#(b(a(x))) → a^#(b(b(x))) (6)

b^#(b(a(x))) → b^#(x) (5)

a^#(a(x)) → b^#(x) (4)

b^#(b(a(x))) → b^#(b(x)) (3)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.