Certification Problem

Input (TPDB TRS_Standard/AG01/#3.15)

The rewrite relation of the following TRS is considered.

average(s(x),y)	→	average(x,s(y))	(1)
average(x,s(s(s(y))))	→	s(average(s(x),y))	(2)
average(0,0)	→	0	(3)
average(0,s(0))	→	0	(4)
average(0,s(s(0)))	→	s(0)	(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.

average^#(s(x),y)	→	average^#(x,s(y))	(6)
average^#(x,s(s(s(y))))	→	average^#(s(x),y)	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

average^#(x,s(s(s(y)))) → average^#(s(x),y) (7)

average^#(s(x),y) → average^#(x,s(y)) (6)

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

[s(x₁)] = x₁ + 1

[average(x₁, x₂)] = 0

[0] = 0

[average^#(x₁, x₂)] = x₁ + x₂ + 0

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