Certification Problem

Input (TPDB TRS_Standard/Rubio_04/lescanne)

The rewrite relation of the following TRS is considered.

div(X,e)	→	i(X)	(1)
i(div(X,Y))	→	div(Y,X)	(2)
div(div(X,Y),Z)	→	div(Y,div(i(X),Z))	(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.

div^#(div(X,Y),Z)	→	div^#(Y,div(i(X),Z))	(4)
div^#(X,e)	→	i^#(X)	(5)
div^#(div(X,Y),Z)	→	div^#(i(X),Z)	(6)
i^#(div(X,Y))	→	div^#(Y,X)	(7)
div^#(div(X,Y),Z)	→	i^#(X)	(8)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

div^#(div(X,Y),Z) → i^#(X) (8)

i^#(div(X,Y)) → div^#(Y,X) (7)

div^#(div(X,Y),Z) → div^#(i(X),Z) (6)

div^#(X,e) → i^#(X) (5)

div^#(div(X,Y),Z) → div^#(Y,div(i(X),Z)) (4)

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

[div^#(x₁, x₂)] = x₁ + 21239

[div(x₁, x₂)] = x₁ + x₂ + 21240

[i(x₁)] = x₁ + 0

[e] = 1

[i^#(x₁)] = x₁ + 0

together with the usable rules

div(X,e) → i(X) (1)

div(div(X,Y),Z) → div(Y,div(i(X),Z)) (3)

i(div(X,Y)) → div(Y,X) (2)

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

div^#(div(X,Y),Z) → i^#(X) (8)

i^#(div(X,Y)) → div^#(Y,X) (7)

div^#(div(X,Y),Z) → div^#(i(X),Z) (6)

div^#(X,e) → i^#(X) (5)

div^#(div(X,Y),Z) → div^#(Y,div(i(X),Z)) (4)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.