Certification Problem

Input (TPDB TRS_Standard/Der95/31)

The rewrite relation of the following TRS is considered.

:(:(x,y),z)	→	:(x,:(y,z))	(1)
:(+(x,y),z)	→	+(:(x,z),:(y,z))	(2)
:(z,+(x,f(y)))	→	:(g(z,y),+(x,a))	(3)

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.

:^#(+(x,y),z)	→	:^#(y,z)	(4)
:^#(:(x,y),z)	→	:^#(y,z)	(5)
:^#(z,+(x,f(y)))	→	:^#(g(z,y),+(x,a))	(6)
:^#(+(x,y),z)	→	:^#(x,z)	(7)
:^#(:(x,y),z)	→	:^#(x,:(y,z))	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

:^#(:(x,y),z) → :^#(x,:(y,z)) (8)

:^#(+(x,y),z) → :^#(x,z) (7)

:^#(:(x,y),z) → :^#(y,z) (5)

:^#(+(x,y),z) → :^#(y,z) (4)

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

[a] = 1

[:^#(x₁, x₂)] = x₁ + 0

[f(x₁)] = 2476

[:(x₁, x₂)] = x₁ + x₂ + 7720

[+(x₁, x₂)] = x₁ + x₂ + 21239

[g(x₁, x₂)] = x₂ + 2476

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pairs

:^#(:(x,y),z) → :^#(x,:(y,z)) (8)

:^#(+(x,y),z) → :^#(x,z) (7)

:^#(:(x,y),z) → :^#(y,z) (5)

:^#(+(x,y),z) → :^#(y,z) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.