Certification Problem

Input (TPDB TRS_Standard/SK90/4.23)

The rewrite relation of the following TRS is considered.

if(true,x,y)	→	x	(1)
if(false,x,y)	→	y	(2)
if(x,y,y)	→	y	(3)
if(if(x,y,z),u,v)	→	if(x,if(y,u,v),if(z,u,v))	(4)
if(x,if(x,y,z),z)	→	if(x,y,z)	(5)
if(x,y,if(x,y,z))	→	if(x,y,z)	(6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the

prec(if)	=	0		stat(if)	=	lex
prec(true)	=	1		stat(true)	=	lex
prec(false)	=	2		stat(false)	=	lex

π(if)	=	[1,3,2]
π(true)	=	[]
π(false)	=	[]

all of the following rules can be deleted.

if(true,x,y)	→	x	(1)
if(false,x,y)	→	y	(2)
if(x,y,y)	→	y	(3)
if(if(x,y,z),u,v)	→	if(x,if(y,u,v),if(z,u,v))	(4)
if(x,if(x,y,z),z)	→	if(x,y,z)	(5)
if(x,y,if(x,y,z))	→	if(x,y,z)	(6)

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.