Certification Problem

Input (TPDB TRS_Standard/SK90/4.22)

The rewrite relation of the following TRS is considered.

not(and(x,y))	→	or(not(x),not(y))	(1)
not(or(x,y))	→	and(not(x),not(y))	(2)
and(x,or(y,z))	→	or(and(x,y),and(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.

not^#(or(x,y))	→	not^#(x)	(4)
not^#(and(x,y))	→	not^#(x)	(5)
not^#(and(x,y))	→	not^#(y)	(6)
and^#(x,or(y,z))	→	and^#(x,z)	(7)
not^#(or(x,y))	→	not^#(y)	(8)
not^#(or(x,y))	→	and^#(not(x),not(y))	(9)
and^#(x,or(y,z))	→	and^#(x,y)	(10)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

not^#(or(x,y)) → not^#(y) (8)

not^#(and(x,y)) → not^#(x) (5)

not^#(and(x,y)) → not^#(y) (6)

not^#(or(x,y)) → not^#(x) (4)

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

[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) (8)

not^#(and(x,y)) → not^#(x) (5)

not^#(and(x,y)) → not^#(y) (6)

not^#(or(x,y)) → not^#(x) (4)

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

and^#(x,or(y,z)) → and^#(x,y) (10)

and^#(x,or(y,z)) → and^#(x,z) (7)

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

[and(x₁, x₂)] = 1

[not^#(x₁)] = 0

[or(x₁, x₂)] = x₁ + x₂ + 1

[and^#(x₁, x₂)] = x₂ + 0

[not(x₁)] = 0

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

and^#(x,or(y,z)) → and^#(x,y) (10)

and^#(x,or(y,z)) → and^#(x,z) (7)

could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.