Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.32)

The rewrite relation of the following TRS is considered.

f(x,c(y))	→	f(x,s(f(y,y)))	(1)
f(s(x),s(y))	→	f(x,s(c(s(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^#(x,c(y))	→	f^#(x,s(f(y,y)))	(3)
f^#(x,c(y))	→	f^#(y,y)	(4)
f^#(s(x),s(y))	→	f^#(x,s(c(s(y))))	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

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

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

[s(x₁)] = 0

[c(x₁)] = x₁ + 1

[f(x₁, x₂)] = 0

[f^#(x₁, x₂)] = x₂ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(x,c(y)) → f^#(y,y) (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,s(c(s(y)))) (5)

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

[s(x₁)] = x₁ + 1

[c(x₁)] = x₁ + 0

[f(x₁, x₂)] = 0

[f^#(x₁, x₂)] = x₁ + 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,s(c(s(y)))) (5)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.