Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/2)

The rewrite relation of the following TRS is considered.

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

→

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

(1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(a,f(f(a,a),x)) → f^#(a,x) (4)

f^#(a,f(f(a,a),x)) → f^#(a,f(a,x)) (3)

1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

[f^#(x₁, x₂)] =

0

0

+

1 1

0 0

· x₁ +

0 1

0 0

· x₂

[a] =

0

0

[f(x₁, x₂)] =

1

0

+

0 0

1 0

· x₁ +

0 0

0 1

· x₂

the pairs

f^#(a,f(f(a,a),x)) → f^#(a,x) (4)

f^#(a,f(f(a,a),x)) → f^#(a,f(a,x)) (3)

could be deleted.
1.1.1.1.1 P is empty

There are no pairs anymore.