Certification Problem

Input (TPDB TRS_Standard/SK90/2.33)

The rewrite relation of the following TRS is considered.

or(x,x)	→	x	(1)
and(x,x)	→	x	(2)
not(not(x))	→	x	(3)
not(and(x,y))	→	or(not(x),not(y))	(4)
not(or(x,y))	→	and(not(x),not(y))	(5)

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.

not^#(or(x,y))	→	not^#(x)	(6)
not^#(and(x,y))	→	not^#(y)	(7)
not^#(and(x,y))	→	not^#(x)	(8)
not^#(or(x,y))	→	not^#(y)	(9)
not^#(or(x,y))	→	and^#(not(x),not(y))	(10)
not^#(and(x,y))	→	or^#(not(x),not(y))	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

not^#(or(x,y)) → not^#(y) (9)

not^#(and(x,y)) → not^#(y) (7)

not^#(and(x,y)) → not^#(x) (8)

not^#(or(x,y)) → not^#(x) (6)

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

[and(x₁, x₂)] = x₁ + x₂ + 1

[not^#(x₁)] = x₁ + 0

[or(x₁, x₂)] = x₁ + x₂ + 1

[or^#(x₁, x₂)] = 0

[and^#(x₁, x₂)] = 0

[not(x₁)] = 0

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

not^#(or(x,y)) → not^#(y) (9)

not^#(and(x,y)) → not^#(y) (7)

not^#(and(x,y)) → not^#(x) (8)

not^#(or(x,y)) → not^#(x) (6)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.