Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z060)

The rewrite relation of the following TRS is considered.

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

→

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

(1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(b(a(b(b(b(x1))))))	→	b^#(b(b(b(b(a(x1))))))	(2)
b^#(b(a(b(b(b(x1))))))	→	b^#(b(b(b(a(x1)))))	(3)
b^#(b(a(b(b(b(x1))))))	→	b^#(b(b(a(x1))))	(4)
b^#(b(a(b(b(b(x1))))))	→	b^#(b(a(x1)))	(5)
b^#(b(a(b(b(b(x1))))))	→	b^#(a(x1))	(6)

1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1

[b(x₁)]

=

x₁ +

1

[a(x₁)]

=

x₁ +

0

[b^#(x₁)]

=

x₁ +

0

together with the usable rule

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

→

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

(1)

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

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

and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
b^#(b(a(b(b(b(x1)))))) → b^#(b(b(b(b(a(x1)))))) (2)

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

[b(x₁)] =

0 1

1 0

· x₁ +

0

0

[a(x₁)] =

0 0

0 0

· x₁ +

1

0

[b^#(x₁)] =

0 3

0 0

· x₁ +

0

0

together with the usable rule
b(b(a(b(b(b(x1)))))) → a(b(b(b(b(b(a(x1))))))) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pair
b^#(b(a(b(b(b(x1)))))) → b^#(b(b(b(b(a(x1)))))) (2)
could be deleted.
1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.