Certification Problem

Input (TPDB TRS_Standard/AProVE_04/improved_usable2)

The rewrite relation of the following TRS is considered.

f(a,x)	→	f(g(x),x)	(1)
h(g(x))	→	h(a)	(2)
g(h(x))	→	g(x)	(3)
h(h(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.

g^#(h(x))	→	g^#(x)	(5)
f^#(a,x)	→	f^#(g(x),x)	(6)
h^#(g(x))	→	h^#(a)	(7)
f^#(a,x)	→	g^#(x)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
f^#(a,x) → f^#(g(x),x) (6)

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

[a] = 2

[h(x₁)] = 0

[f(x₁, x₂)] = 0

[h^#(x₁)] = 0

[f^#(x₁, x₂)] = x₁ + 0

[g^#(x₁)] = 0

[g(x₁)] = 1

together with the usable rule
g(h(x)) → g(x) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
f^#(a,x) → f^#(g(x),x) (6)
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^#(h(x)) → g^#(x) (5)

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

[a] = 2

[h(x₁)] = x₁ + 1

[f(x₁, x₂)] = 0

[h^#(x₁)] = 0

[f^#(x₁, x₂)] = x₁ + 0

[g^#(x₁)] = x₁ + 0

[g(x₁)] = 1

together with the usable rule
g(h(x)) → g(x) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
g^#(h(x)) → g^#(x) (5)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.