Certification Problem

Input (TPDB TRS_Standard/Zantema_05/jw37)

The rewrite relation of the following TRS is considered.

f(f(a,x),a)

→

f(f(f(x,a),f(a,a)),a)

(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.

f^#(f(a,x),a)	→	f^#(f(x,a),f(a,a))	(2)
f^#(f(a,x),a)	→	f^#(x,a)	(3)
f^#(f(a,x),a)	→	f^#(f(f(x,a),f(a,a)),a)	(4)
f^#(f(a,x),a)	→	f^#(a,a)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(f(a,x),a) → f^#(f(f(x,a),f(a,a)),a) (4)

f^#(f(a,x),a) → f^#(x,a) (3)

1.1.1 Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals

[a] =

1

0

[f(x₁, x₂)] =

0 0

1 0

· x₁ +

0 0

0 1

· x₂ +

0

0

[f^#(x₁, x₂)] =

0 1

0 0

· x₁ +

0

0

together with the usable rule
f(f(a,x),a) → f(f(f(x,a),f(a,a)),a) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(f(a,x),a) → f^#(x,a) (3)
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^#(f(a,x),a) → f^#(f(f(x,a),f(a,a)),a) (4)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
  
  [a] =
  
  1
  
  1
  
  [f(x₁, x₂)] =
  
  0 0
  
  1 0
  
  · x₁ +
  
  0
  
  1
  
  [f^#(x₁, x₂)] =
  
  0 1
  
  0 1
  
  · x₁ +
  
  0
  
  0
  
  together with the usable rule
  f(f(a,x),a) → f(f(f(x,a),f(a,a)),a) (1)
  (w.r.t. the implicit argument filter of the reduction pair), the pair
  f^#(f(a,x),a) → f^#(f(f(x,a),f(a,a)),a) (4)
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.