Certification Problem

Input (TPDB TRS_Standard/SK90/2.21)

The rewrite relation of the following TRS is considered.

bin(x,0)	→	s(0)	(1)
bin(0,s(y))	→	0	(2)
bin(s(x),s(y))	→	+(bin(x,s(y)),bin(x,y))	(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.

bin^#(s(x),s(y))	→	bin^#(x,y)	(4)
bin^#(s(x),s(y))	→	bin^#(x,s(y))	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

bin^#(s(x),s(y)) → bin^#(x,s(y)) (5)

bin^#(s(x),s(y)) → bin^#(x,y) (4)

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

[s(x₁)] = x₁ + 1

[bin^#(x₁, x₂)] = x₁ + 0

[0] = 0

[bin(x₁, x₂)] = 0

[+(x₁, x₂)] = 0

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

bin^#(s(x),s(y)) → bin^#(x,s(y)) (5)

bin^#(s(x),s(y)) → bin^#(x,y) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.