Certification Problem

Input (TPDB TRS_Standard/SK90/2.03)

The rewrite relation of the following TRS is considered.

minus(minus(x))	→	x	(1)
minus(h(x))	→	h(minus(x))	(2)
minus(f(x,y))	→	f(minus(y),minus(x))	(3)

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.

minus^#(f(x,y))	→	minus^#(y)	(4)
minus^#(f(x,y))	→	minus^#(x)	(5)
minus^#(h(x))	→	minus^#(x)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

minus^#(h(x)) → minus^#(x) (6)

minus^#(f(x,y)) → minus^#(x) (5)

minus^#(f(x,y)) → minus^#(y) (4)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[h(x₁)] = x₁ + 1

[minus(x₁)] = 0

[f(x₁, x₂)] = x₁ + x₂ + 1

[minus^#(x₁)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

minus^#(h(x)) → minus^#(x) (6)

minus^#(f(x,y)) → minus^#(x) (5)

minus^#(f(x,y)) → minus^#(y) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.