Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/jones5)

The rewrite relation of the following TRS is considered.

f(x,empty)	→	x	(1)
f(empty,cons(a,k))	→	f(cons(a,k),k)	(2)
f(cons(a,k),y)	→	f(y,k)	(3)

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^#(cons(a,k),y)	→	f^#(y,k)	(4)
f^#(empty,cons(a,k))	→	f^#(cons(a,k),k)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(empty,cons(a,k)) → f^#(cons(a,k),k) (5)

f^#(cons(a,k),y) → f^#(y,k) (4)

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

[f(x₁, x₂)] = max(0)

[f^#(x₁, x₂)] = max(x₁ + 11293, x₂ + 32532, 0)

[cons(x₁, x₂)] = max(x₂ + 21239, 0)

[empty] = 21240

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(empty,cons(a,k)) → f^#(cons(a,k),k) (5)
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^#(cons(a,k),y) → f^#(y,k) (4)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the Max-polynomial interpretation
  
  [f(x₁, x₂)] = 0
  
  [f^#(x₁, x₂)] = x₁ + x₂ + 0
  
  [cons(x₁, x₂)] = x₂ + 1
  
  [empty] = 0
  
  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
  f^#(cons(a,k),y) → f^#(y,k) (4)
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.