Certification Problem

Input (TPDB TRS_Standard/SK90/4.05)

The rewrite relation of the following TRS is considered.

*(x,+(y,z))	→	+((x,y),(x,z))	(1)
*(+(x,y),z)	→	+((x,z),(y,z))	(2)
*(x,1)	→	x	(3)
*(1,y)	→	y	(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,+(y,z))	→	*^#(x,y)	(6)
*^#(+(x,y),z)	→	*^#(x,z)	(7)
*^#(x,+(y,z))	→	*^#(x,z)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

*^#(x,+(y,z)) → *^#(x,z) (8)

*^#(+(x,y),z) → *^#(x,z) (7)

*^#(x,+(y,z)) → *^#(x,y) (6)

*^#(+(x,y),z) → *^#(y,z) (5)

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

[1] = 0

[*^#(x₁, x₂)] = x₂ + 0

[+(x₁, x₂)] = x₁ + x₂ + 1

[*(x₁, x₂)] = 0

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

*^#(x,+(y,z)) → *^#(x,z) (8)

*^#(x,+(y,z)) → *^#(x,y) (6)

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,y),z) → *^#(y,z) (5)
  
  *^#(+(x,y),z) → *^#(x,z) (7)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the Max-polynomial interpretation
  
  [1] = 0
  
  [*^#(x₁, x₂)] = x₁ + 0
  
  [+(x₁, x₂)] = x₁ + x₂ + 1
  
  [*(x₁, x₂)] = 0
  
  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs
  
  *^#(+(x,y),z) → *^#(y,z) (5)
  
  *^#(+(x,y),z) → *^#(x,z) (7)
  
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.