Certification Problem

Input (TPDB TRS_Standard/Various_04/22)

The rewrite relation of the following TRS is considered.

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

g^#(0,x)	→	g^#(f(x,x),x)	(4)
g^#(0,x)	→	f^#(x,x)	(5)
f^#(s(x),s(y))	→	f^#(x,y)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
g^#(0,x) → g^#(f(x,x),x) (4)

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

[s(x₁)] = 1

[f(x₁, x₂)] = 1

[0] = 2

[f^#(x₁, x₂)] = 0

[g^#(x₁, x₂)] = x₁ + 0

[g(x₁, x₂)] = 0

together with the usable rules

f(x,0) → s(0) (1)

f(s(x),s(y)) → s(f(x,y)) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
g^#(0,x) → g^#(f(x,x),x) (4)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.
The 2^nd component contains the pair
f^#(s(x),s(y)) → f^#(x,y) (6)

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

[s(x₁)] = x₁ + 1

[f(x₁, x₂)] = 21239

[0] = 0

[f^#(x₁, x₂)] = x₁ + 0

[g^#(x₁, x₂)] = 0

[g(x₁, x₂)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(s(x),s(y)) → f^#(x,y) (6)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.