Certification Problem

Input (TPDB TRS_Standard/AG01/#3.23)

The rewrite relation of the following TRS is considered.

f(0,y)	→	0	(1)
f(s(x),y)	→	f(f(x,y),y)	(2)

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),y)	→	f^#(x,y)	(3)
f^#(s(x),y)	→	f^#(f(x,y),y)	(4)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(s(x),y) → f^#(f(x,y),y) (4)

f^#(s(x),y) → f^#(x,y) (3)

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

[s(x₁)] = x₁ + 2

[f(x₁, x₂)] = x₁ + 1

[0] = 28958

[f^#(x₁, x₂)] = x₁ + 0

together with the usable rules

f(0,y) → 0 (1)

f(s(x),y) → f(f(x,y),y) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pairs

f^#(s(x),y) → f^#(f(x,y),y) (4)

f^#(s(x),y) → f^#(x,y) (3)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.