Certification Problem

Input (TPDB TRS_Standard/SK90/4.20)

The rewrite relation of the following TRS is considered.

not(x)	→	xor(x,true)	(1)
or(x,y)	→	xor(and(x,y),xor(x,y))	(2)
implies(x,y)	→	xor(and(x,y),xor(x,true))	(3)
and(x,true)	→	x	(4)
and(x,false)	→	false	(5)
and(x,x)	→	x	(6)
xor(x,false)	→	x	(7)
xor(x,x)	→	false	(8)
and(xor(x,y),z)	→	xor(and(x,z),and(y,z))	(9)

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.

or^#(x,y)	→	and^#(x,y)	(10)
and^#(xor(x,y),z)	→	and^#(y,z)	(11)
implies^#(x,y)	→	xor^#(x,true)	(12)
implies^#(x,y)	→	xor^#(and(x,y),xor(x,true))	(13)
not^#(x)	→	xor^#(x,true)	(14)
and^#(xor(x,y),z)	→	and^#(x,z)	(15)
or^#(x,y)	→	xor^#(x,y)	(16)
or^#(x,y)	→	xor^#(and(x,y),xor(x,y))	(17)
implies^#(x,y)	→	and^#(x,y)	(18)
and^#(xor(x,y),z)	→	xor^#(and(x,z),and(y,z))	(19)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

and^#(xor(x,y),z)	→	and^#(y,z)	(11)
and^#(xor(x,y),z)	→	and^#(x,z)	(15)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[and(x₁, x₂)]	=	0
[false]	=	0
[true]	=	0
[not^#(x₁)]	=	0
[implies^#(x₁, x₂)]	=	0
[or(x₁, x₂)]	=	0
[implies(x₁, x₂)]	=	0
[xor^#(x₁, x₂)]	=	0
[or^#(x₁, x₂)]	=	0
[xor(x₁, x₂)]	=	x₁ + x₂ + 1
[and^#(x₁, x₂)]	=	x₁ + 0
[not(x₁)]	=	0

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

and^#(xor(x,y),z)	→	and^#(y,z)	(11)
and^#(xor(x,y),z)	→	and^#(x,z)	(15)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.