Certification Problem

Input (TPDB TRS_Standard/Various_04/18)

The rewrite relation of the following TRS is considered.

+((x,y),(x,z))	→	*(x,+(y,z))	(1)
+(+(x,y),z)	→	+(x,+(y,z))	(2)
+((x,y),+((x,z),u))	→	+(*(x,+(y,z)),u)	(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.

+^#(+(x,y),z)	→	+^#(y,z)	(4)
+^#((x,y),(x,z))	→	+^#(y,z)	(5)
+^#((x,y),+((x,z),u))	→	+^#(*(x,+(y,z)),u)	(6)
+^#(+(x,y),z)	→	+^#(x,+(y,z))	(7)
+^#((x,y),+((x,z),u))	→	+^#(y,z)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

+^#(*(x,y),+(*(x,z),u)) → +^#(y,z) (8)

+^#(+(x,y),z) → +^#(x,+(y,z)) (7)

+^#(*(x,y),*(x,z)) → +^#(y,z) (5)

+^#(+(x,y),z) → +^#(y,z) (4)

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

[u] = 1

[+(x₁, x₂)] = x₁ + x₂ + 1

[+^#(x₁, x₂)] = x₁ + 0

[*(x₁, x₂)] = x₁ + x₂ + 39

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

+^#(*(x,y),+(*(x,z),u)) → +^#(y,z) (8)

+^#(+(x,y),z) → +^#(x,+(y,z)) (7)

+^#(*(x,y),*(x,z)) → +^#(y,z) (5)

+^#(+(x,y),z) → +^#(y,z) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.