Certification Problem

Input (TPDB TRS_Standard/SK90/4.10)

The rewrite relation of the following TRS is considered.

(x,(y,z))	→	*(otimes(x,y),z)	(1)
*(1,y)	→	y	(2)
*(+(x,y),z)	→	oplus((x,z),(y,z))	(3)
*(x,oplus(y,z))	→	oplus((x,y),(x,z))	(4)

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)	(5)
*^#(x,oplus(y,z))	→	*^#(x,z)	(6)
^#(x,(y,z))	→	*^#(otimes(x,y),z)	(7)
*^#(+(x,y),z)	→	*^#(x,z)	(8)
*^#(x,oplus(y,z))	→	*^#(x,y)	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

*^#(x,oplus(y,z)) → *^#(x,y) (9)

*^#(+(x,y),z) → *^#(x,z) (8)

*^#(x,*(y,z)) → *^#(otimes(x,y),z) (7)

*^#(x,oplus(y,z)) → *^#(x,z) (6)

*^#(+(x,y),z) → *^#(y,z) (5)

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

[oplus(x₁, x₂)] = x₁ + x₂ + 1

[1] = 0

[*^#(x₁, x₂)] = x₁ + x₂ + 0

[otimes(x₁, x₂)] = 1

[+(x₁, x₂)] = x₁ + x₂ + 1

[*(x₁, x₂)] = x₁ + x₂ + 2

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

*^#(x,oplus(y,z)) → *^#(x,y) (9)

*^#(+(x,y),z) → *^#(x,z) (8)

*^#(x,*(y,z)) → *^#(otimes(x,y),z) (7)

*^#(x,oplus(y,z)) → *^#(x,z) (6)

*^#(+(x,y),z) → *^#(y,z) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.