Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z126)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
a^#(b(a(x1))) → a^#(b(b(a(x1)))) (4)

1.1.1 Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

[a(x₁)] =

1 1

1 0

· x₁ +

1

2

[b(x₁)] =

0 1

0 0

· x₁ +

1

1

[a^#(x₁)] =

1 1

0 0

· x₁ +

0

0

[b^#(x₁)] =

0

0

together with the usable rules

a(b(a(x1))) → a(b(b(a(x1)))) (1)

b(b(b(x1))) → b(b(x1)) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
a^#(b(a(x1))) → a^#(b(b(a(x1)))) (4)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.