Certification Problem

Input (TPDB TRS_Standard/SK90/4.04)

The rewrite relation of the following TRS is considered.

+((x,y),(a,y))	→	*(+(x,a),y)	(1)
((x,y),z)	→	(x,(y,z))	(2)

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.

^#((x,y),z)	→	*^#(y,z)	(3)
+^#((x,y),(a,y))	→	*^#(+(x,a),y)	(4)
^#((x,y),z)	→	^#(x,(y,z))	(5)
+^#((x,y),(a,y))	→	+^#(x,a)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

*^#(*(x,y),z) → *^#(x,*(y,z)) (5)

*^#(*(x,y),z) → *^#(y,z) (3)

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

[a] = 0

[*^#(x₁, x₂)] = x₁ + 0

[+(x₁, x₂)] = 0

[+^#(x₁, x₂)] = 0

[*(x₁, x₂)] = x₁ + x₂ + 1

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

*^#(*(x,y),z) → *^#(x,*(y,z)) (5)

*^#(*(x,y),z) → *^#(y,z) (3)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.