Certification Problem

Input (TPDB TRS_Standard/SK90/2.36)

The rewrite relation of the following TRS is considered.

implies(not(x),y)	→	or(x,y)	(1)
implies(not(x),or(y,z))	→	implies(y,or(x,z))	(2)
implies(x,or(y,z))	→	or(y,implies(x,z))	(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.

implies^#(x,or(y,z))	→	implies^#(x,z)	(4)
implies^#(not(x),or(y,z))	→	implies^#(y,or(x,z))	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

implies^#(not(x),or(y,z)) → implies^#(y,or(x,z)) (5)

implies^#(x,or(y,z)) → implies^#(x,z) (4)

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

[implies^#(x₁, x₂)] = x₁ + x₂ + 0

[or(x₁, x₂)] = x₁ + x₂ + 1

[implies(x₁, x₂)] = 0

[not(x₁)] = x₁ + 1

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

implies^#(not(x),or(y,z)) → implies^#(y,or(x,z)) (5)

implies^#(x,or(y,z)) → implies^#(x,z) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.