Certification Problem

Input (TPDB TRS_Standard/AG01/#3.53b)

The rewrite relation of the following TRS is considered.

g(x,y)	→	x	(1)
g(x,y)	→	y	(2)
f(0,1,x)	→	f(s(x),x,x)	(3)
f(x,y,s(z))	→	s(f(0,1,z))	(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^#(x,y,s(z))	→	f^#(0,1,z)	(5)
f^#(0,1,x)	→	f^#(s(x),x,x)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(0,1,x) → f^#(s(x),x,x) (6)

f^#(x,y,s(z)) → f^#(0,1,z) (5)

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

[1] = 1

[s(x₁)] = x₁ + 1

[f(x₁, x₂, x₃)] = 0

[0] = 1

[f^#(x₁, x₂, x₃)] = x₃ + 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^#(x,y,s(z)) → f^#(0,1,z) (5)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.