Certification Problem

Input (TPDB TRS_Standard/Various_04/26)

The rewrite relation of the following TRS is considered.

f(f(x))	→	f(g(f(x),x))	(1)
f(f(x))	→	f(h(f(x),f(x)))	(2)
g(x,y)	→	y	(3)
h(x,x)	→	g(x,0)	(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))	→	h^#(f(x),f(x))	(5)
h^#(x,x)	→	g^#(x,0)	(6)
f^#(f(x))	→	f^#(g(f(x),x))	(7)
f^#(f(x))	→	f^#(h(f(x),f(x)))	(8)
f^#(f(x))	→	g^#(f(x),x)	(9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(f(x)) → f^#(h(f(x),f(x))) (8)

f^#(f(x)) → f^#(g(f(x),x)) (7)

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

[h(x₁, x₂)] = 23677

[f(x₁)] = x₁ + 23678

[0] = 1

[h^#(x₁, x₂)] = 0

[f^#(x₁)] = x₁ + 0

[g^#(x₁, x₂)] = 0

[g(x₁, x₂)] = x₂ + 23676

together with the usable rules

h(x,x) → g(x,0) (4)

f(f(x)) → f(g(f(x),x)) (1)

g(x,y) → y (3)

f(f(x)) → f(h(f(x),f(x))) (2)

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

f^#(f(x)) → f^#(h(f(x),f(x))) (8)

f^#(f(x)) → f^#(g(f(x),x)) (7)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.