Certification Problem

Input (TPDB TRS_Standard/AProVE_04/IJCAR_1)

The rewrite relation of the following TRS is considered.

div(0,y)	→	0	(1)
div(x,y)	→	quot(x,y,y)	(2)
quot(0,s(y),z)	→	0	(3)
quot(s(x),s(y),z)	→	quot(x,y,z)	(4)
quot(x,0,s(z))	→	s(div(x,s(z)))	(5)

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.

quot^#(x,0,s(z))	→	div^#(x,s(z))	(6)
quot^#(s(x),s(y),z)	→	quot^#(x,y,z)	(7)
div^#(x,y)	→	quot^#(x,y,y)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

div^#(x,y) → quot^#(x,y,y) (8)

quot^#(s(x),s(y),z) → quot^#(x,y,z) (7)

quot^#(x,0,s(z)) → div^#(x,s(z)) (6)

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

[div^#(x₁, x₂)] = x₁ + 0

[s(x₁)] = x₁ + 1

[div(x₁, x₂)] = 0

[0] = 1

[quot(x₁, x₂, x₃)] = 0

[quot^#(x₁, x₂, x₃)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
quot^#(s(x),s(y),z) → quot^#(x,y,z) (7)
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
  
  quot^#(x,0,s(z)) → div^#(x,s(z)) (6)
  
  div^#(x,y) → quot^#(x,y,y) (8)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the Max-polynomial interpretation
  
  [div^#(x₁, x₂)] = x₂ + 1
  
  [s(x₁)] = 1
  
  [div(x₁, x₂)] = 0
  
  [0] = 3
  
  [quot(x₁, x₂, x₃)] = 0
  
  [quot^#(x₁, x₂, x₃)] = x₂ + 0
  
  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs
  
  quot^#(x,0,s(z)) → div^#(x,s(z)) (6)
  
  div^#(x,y) → quot^#(x,y,y) (8)
  
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.