Certification Problem

Input (TPDB TRS_Standard/SK90/4.39)

The rewrite relation of the following TRS is considered.

*(x,*(minus(y),y))

→

*(minus(*(y,y)),x)

(1)

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,(minus(y),y))	→	*^#(y,y)	(2)
^#(x,(minus(y),y))	→	^#(minus((y,y)),x)	(3)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

*^#(x,*(minus(y),y)) → *^#(minus(*(y,y)),x) (3)

*^#(x,*(minus(y),y)) → *^#(y,y) (2)

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

[minus(x₁)] = x₁ + 20163

[*^#(x₁, x₂)] = x₁ + x₂ + 0

[*(x₁, x₂)] = x₁ + x₂ + 0

together with the usable rule
*(x,*(minus(y),y)) → *(minus(*(y,y)),x) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pair
*^#(x,*(minus(y),y)) → *^#(y,y) (2)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.
- The 1^st component contains the pair
  *^#(x,*(minus(y),y)) → *^#(minus(*(y,y)),x) (3)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the Max-polynomial interpretation
  
  [minus(x₁)] = 1
  
  [*^#(x₁, x₂)] = x₁ + x₂ + 0
  
  [*(x₁, x₂)] = x₂ + 2
  
  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
  *^#(x,*(minus(y),y)) → *^#(minus(*(y,y)),x) (3)
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.