Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/matchbox2)

The rewrite relation of the following TRS is considered.

f(f(x,a),y)

→

f(f(a,y),f(a,x))

(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.

f^#(f(x,a),y)	→	f^#(a,y)	(2)
f^#(f(x,a),y)	→	f^#(a,x)	(3)
f^#(f(x,a),y)	→	f^#(f(a,y),f(a,x))	(4)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
f^#(f(x,a),y) → f^#(f(a,y),f(a,x)) (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] =

1

28473

[f(x₁, x₂)] =

1 0

0 0

· x₁ +

0 1

0 0

· x₂ +

3019

28471

[f^#(x₁, x₂)] =

1 0

0 0

· x₁ +

0 1

0 0

· x₂ +

0

0

together with the usable rule
f(f(x,a),y) → f(f(a,y),f(a,x)) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(f(x,a),y) → f^#(f(a,y),f(a,x)) (4)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.