Certification Problem

Input (TPDB SRS_Standard/Zantema_06/02)

The rewrite relation of the following TRS is considered.

a(p(x1))	→	p(a(A(x1)))	(1)
a(A(x1))	→	A(a(x1))	(2)
p(A(A(x1)))	→	a(p(x1))	(3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

p(a(x1))	→	A(a(p(x1)))	(4)
A(a(x1))	→	a(A(x1))	(5)
A(A(p(x1)))	→	p(a(x1))	(6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

A^#(A(p(x1)))	→	p^#(a(x1))	(7)
A^#(a(x1))	→	A^#(x1)	(8)
p^#(a(x1))	→	A^#(a(p(x1)))	(9)
p^#(a(x1))	→	p^#(x1)	(10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

A^#(A(p(x1))) → p^#(a(x1)) (7)

p^#(a(x1)) → A^#(a(p(x1))) (9)

A^#(a(x1)) → A^#(x1) (8)

p^#(a(x1)) → p^#(x1) (10)

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

[A(x₁)] =

1

· x₁ +

1

[a(x₁)] =

1

· x₁ +

1

[p(x₁)] =

2

· x₁ +

1

[A^#(x₁)] =

4

· x₁ +

3

[p^#(x₁)] =

8

· x₁ +

3

together with the usable rules

p(a(x1)) → A(a(p(x1))) (4)

A(a(x1)) → a(A(x1)) (5)

A(A(p(x1))) → p(a(x1)) (6)

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

A^#(a(x1)) → A^#(x1) (8)

p^#(a(x1)) → p^#(x1) (10)

and no rules could be deleted.
1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.