Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z033)

The rewrite relation of the following TRS is considered.

b(a(b(b(a(b(a(x1)))))))

→

a(b(a(a(b(b(a(b(b(a(x1))))))))))

(1)

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(b(b(a(b(a(x1)))))))	→	b^#(b(a(b(b(a(x1))))))	(2)
b^#(a(b(b(a(b(a(x1)))))))	→	b^#(a(b(b(a(x1)))))	(3)
b^#(a(b(b(a(b(a(x1)))))))	→	b^#(a(a(b(b(a(b(b(a(x1)))))))))	(4)
b^#(a(b(b(a(b(a(x1)))))))	→	b^#(b(a(x1)))	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

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

b^#(a(b(b(a(b(a(x1))))))) → b^#(b(a(b(b(a(x1)))))) (2)

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)] =

0 0

1 0

· x₁ +

1

0

[b(x₁)] =

1 1

0 1

· x₁ +

0

1

[b^#(x₁)] =

0 1

0 0

· x₁ +

0

0

together with the usable rule
b(a(b(b(a(b(a(x1))))))) → a(b(a(a(b(b(a(b(b(a(x1)))))))))) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pairs

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

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

b^#(a(b(b(a(b(a(x1))))))) → b^#(b(a(b(b(a(x1)))))) (2)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.