Certification Problem

Input (TPDB TRS_Standard/Zantema_05/jw42)

The rewrite relation of the following TRS is considered.

f(f(a,x),a)

→

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

(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(a,x),a)	→	f^#(f(a,f(a,a)),x)	(2)
f^#(f(a,x),a)	→	f^#(a,f(a,a))	(3)
f^#(f(a,x),a)	→	f^#(f(f(a,f(a,a)),x),a)	(4)
f^#(f(a,x),a)	→	f^#(a,a)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(f(a,x),a) → f^#(f(f(a,f(a,a)),x),a) (4)

f^#(f(a,x),a) → f^#(f(a,f(a,a)),x) (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] =

2

1

[f(x₁, x₂)] =

0 0

1 0

· x₁ +

0 0

1 0

· x₂ +

1

0

[f^#(x₁, x₂)] =

0 1

1 1

· x₁ +

1 0

1 0

· x₂ +

0

0

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

f^#(f(a,x),a) → f^#(f(f(a,f(a,a)),x),a) (4)

f^#(f(a,x),a) → f^#(f(a,f(a,a)),x) (2)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.