Certification Problem

Input (TPDB TRS_Standard/SK90/2.07)

The rewrite relation of the following TRS is considered.

f(0,y)	→	y	(1)
f(x,0)	→	x	(2)
f(i(x),y)	→	i(x)	(3)
f(f(x,y),z)	→	f(x,f(y,z))	(4)
f(g(x,y),z)	→	g(f(x,z),f(y,z))	(5)
f(1,g(x,y))	→	x	(6)
f(2,g(x,y))	→	y	(7)

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^#(g(x,y),z)	→	f^#(y,z)	(8)
f^#(f(x,y),z)	→	f^#(x,f(y,z))	(9)
f^#(g(x,y),z)	→	f^#(x,z)	(10)
f^#(f(x,y),z)	→	f^#(y,z)	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

f^#(f(x,y),z) → f^#(y,z) (11)

f^#(g(x,y),z) → f^#(x,z) (10)

f^#(f(x,y),z) → f^#(x,f(y,z)) (9)

f^#(g(x,y),z) → f^#(y,z) (8)

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

[1] = 1

[f(x₁, x₂)] = x₁ + x₂ + 1

[0] = 40005

[f^#(x₁, x₂)] = x₁ + 0

[i(x₁)] = 33036

[2] = 10803

[g(x₁, x₂)] = x₁ + x₂ + 8856

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

f^#(f(x,y),z) → f^#(y,z) (11)

f^#(g(x,y),z) → f^#(x,z) (10)

f^#(f(x,y),z) → f^#(x,f(y,z)) (9)

f^#(g(x,y),z) → f^#(y,z) (8)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.