Certification Problem

Input (TPDB TRS_Standard/SK90/2.32)

The rewrite relation of the following TRS is considered.

not(x)	→	if(x,false,true)	(1)
and(x,y)	→	if(x,y,false)	(2)
or(x,y)	→	if(x,true,y)	(3)
implies(x,y)	→	if(x,y,true)	(4)
=(x,x)	→	true	(5)
=(x,y)	→	if(x,y,not(y))	(6)
if(true,x,y)	→	x	(7)
if(false,x,y)	→	y	(8)
if(x,x,if(x,false,true))	→	true	(9)
=(x,y)	→	if(x,y,if(y,false,true))	(10)

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.

=^#(x,y)	→	if^#(x,y,not(y))	(11)
or^#(x,y)	→	if^#(x,true,y)	(12)
not^#(x)	→	if^#(x,false,true)	(13)
=^#(x,y)	→	if^#(y,false,true)	(14)
=^#(x,y)	→	not^#(y)	(15)
and^#(x,y)	→	if^#(x,y,false)	(16)
implies^#(x,y)	→	if^#(x,y,true)	(17)
=^#(x,y)	→	if^#(x,y,if(y,false,true))	(18)

1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.