Certification Problem

Input (TPDB TRS_Standard/Rubio_04/test4)

The rewrite relation of the following TRS is considered.

f(a,a)	→	f(a,b)	(1)
f(a,b)	→	f(s(a),c)	(2)
f(s(X),c)	→	f(X,c)	(3)
f(c,c)	→	f(a,a)	(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.

f^#(s(X),c)	→	f^#(X,c)	(5)
f^#(a,a)	→	f^#(a,b)	(6)
f^#(a,b)	→	f^#(s(a),c)	(7)
f^#(c,c)	→	f^#(a,a)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(c,c) → f^#(a,a) (8)

f^#(a,b) → f^#(s(a),c) (7)

f^#(a,a) → f^#(a,b) (6)

f^#(s(X),c) → f^#(X,c) (5)

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

[a] = 40651

[s(x₁)] = x₁ + 0

[b] = 1

[c] = 40652

[f(x₁, x₂)] = 0

[f^#(x₁, x₂)] = x₁ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(c,c) → f^#(a,a) (8)
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
  f^#(s(X),c) → f^#(X,c) (5)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the Max-polynomial interpretation
  
  [a] = 40651
  
  [s(x₁)] = x₁ + 1
  
  [b] = 1
  
  [c] = 40652
  
  [f(x₁, x₂)] = 0
  
  [f^#(x₁, x₂)] = x₁ + 0
  
  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
  f^#(s(X),c) → f^#(X,c) (5)
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.