Certification Problem

Input (TPDB TRS_Standard/AG01/#3.56)

The rewrite relation of the following TRS is considered.

g(c(x,s(y)))	→	g(c(s(x),y))	(1)
f(c(s(x),y))	→	f(c(x,s(y)))	(2)
f(f(x))	→	f(d(f(x)))	(3)
f(x)	→	x	(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^#(f(x))	→	f^#(d(f(x)))	(5)
g^#(c(x,s(y)))	→	g^#(c(s(x),y))	(6)
f^#(c(s(x),y))	→	f^#(c(x,s(y)))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
f^#(c(s(x),y)) → f^#(c(x,s(y))) (7)

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

[d(x₁)] = 0

[s(x₁)] = x₁ + 1

[c(x₁, x₂)] = x₁ + 28958

[f(x₁)] = 0

[f^#(x₁)] = x₁ + 0

[g^#(x₁)] = 0

[g(x₁)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(c(s(x),y)) → f^#(c(x,s(y))) (7)
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
g^#(c(x,s(y))) → g^#(c(s(x),y)) (6)

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

[d(x₁)] = 0

[s(x₁)] = x₁ + 1

[c(x₁, x₂)] = x₂ + 40651

[f(x₁)] = 0

[f^#(x₁)] = 0

[g^#(x₁)] = x₁ + 0

[g(x₁)] = 0

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

The dependency pairs are split into 0 components.