Certification Problem

Input (TPDB TRS_Standard/Rubio_04/bn122)

The rewrite relation of the following TRS is considered.

plus(plus(X,Y),Z)	→	plus(X,plus(Y,Z))	(1)
times(X,s(Y))	→	plus(X,times(Y,X))	(2)

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.

times^#(X,s(Y))	→	times^#(Y,X)	(3)
plus^#(plus(X,Y),Z)	→	plus^#(X,plus(Y,Z))	(4)
times^#(X,s(Y))	→	plus^#(X,times(Y,X))	(5)
plus^#(plus(X,Y),Z)	→	plus^#(Y,Z)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
times^#(X,s(Y)) → times^#(Y,X) (3)

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

[s(x₁)] = x₁ + 1

[plus^#(x₁, x₂)] = 0

[times^#(x₁, x₂)] = x₁ + x₂ + 0

[times(x₁, x₂)] = 0

[plus(x₁, x₂)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
times^#(X,s(Y)) → times^#(Y,X) (3)
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

plus^#(plus(X,Y),Z) → plus^#(Y,Z) (6)

plus^#(plus(X,Y),Z) → plus^#(X,plus(Y,Z)) (4)

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

[s(x₁)] = 1

[plus^#(x₁, x₂)] = x₁ + 0

[times^#(x₁, x₂)] = 0

[times(x₁, x₂)] = 0

[plus(x₁, x₂)] = x₁ + x₂ + 1

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

plus^#(plus(X,Y),Z) → plus^#(Y,Z) (6)

plus^#(plus(X,Y),Z) → plus^#(X,plus(Y,Z)) (4)

could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.