Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_CSR_05/ExAppendixB_AEL03)

The rewrite relation of the following TRS is considered.

from(X)	→	cons(X,from(s(X)))	(1)
2ndspos(0,Z)	→	rnil	(2)
2ndspos(s(N),cons(X,Z))	→	2ndspos(s(N),cons2(X,Z))	(3)
2ndspos(s(N),cons2(X,cons(Y,Z)))	→	rcons(posrecip(Y),2ndsneg(N,Z))	(4)
2ndsneg(0,Z)	→	rnil	(5)
2ndsneg(s(N),cons(X,Z))	→	2ndsneg(s(N),cons2(X,Z))	(6)
2ndsneg(s(N),cons2(X,cons(Y,Z)))	→	rcons(negrecip(Y),2ndspos(N,Z))	(7)
pi(X)	→	2ndspos(X,from(0))	(8)
plus(0,Y)	→	Y	(9)
plus(s(X),Y)	→	s(plus(X,Y))	(10)
times(0,Y)	→	0	(11)
times(s(X),Y)	→	plus(Y,times(X,Y))	(12)
square(X)	→	times(X,X)	(13)

Property / Task

Prove or disprove termination.

Answer / Result

No.

Proof (by AProVE @ termCOMP 2023)

1 Innermost Lhss Increase

We add the following left hand sides to the innermost strategy.

from(x0)

2ndspos(0,x0)

2ndspos(s(x0),cons(x1,x2))

2ndspos(s(x0),cons2(x1,cons(x2,x3)))

2ndsneg(0,x0)

2ndsneg(s(x0),cons(x1,x2))

2ndsneg(s(x0),cons2(x1,cons(x2,x3)))

pi(x0)

plus(0,x0)

plus(s(x0),x1)

times(0,x0)

times(s(x0),x1)

square(x0)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

from^#(X)	→	from^#(s(X))	(14)
2ndspos^#(s(N),cons(X,Z))	→	2ndspos^#(s(N),cons2(X,Z))	(15)
2ndspos^#(s(N),cons2(X,cons(Y,Z)))	→	2ndsneg^#(N,Z)	(16)
2ndsneg^#(s(N),cons(X,Z))	→	2ndsneg^#(s(N),cons2(X,Z))	(17)
2ndsneg^#(s(N),cons2(X,cons(Y,Z)))	→	2ndspos^#(N,Z)	(18)
pi^#(X)	→	2ndspos^#(X,from(0))	(19)
pi^#(X)	→	from^#(0)	(20)
plus^#(s(X),Y)	→	plus^#(X,Y)	(21)
times^#(s(X),Y)	→	plus^#(Y,times(X,Y))	(22)
times^#(s(X),Y)	→	times^#(X,Y)	(23)
square^#(X)	→	times^#(X,X)	(24)

It remains to prove infiniteness of the resulting DP problem.

1.1.1 Pair and Rule Removal

Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem. The following pairs have been deleted.

2ndspos^#(s(N),cons(X,Z))	→	2ndspos^#(s(N),cons2(X,Z))	(15)
2ndspos^#(s(N),cons2(X,cons(Y,Z)))	→	2ndsneg^#(N,Z)	(16)
2ndsneg^#(s(N),cons(X,Z))	→	2ndsneg^#(s(N),cons2(X,Z))	(17)
2ndsneg^#(s(N),cons2(X,cons(Y,Z)))	→	2ndspos^#(N,Z)	(18)
pi^#(X)	→	2ndspos^#(X,from(0))	(19)
pi^#(X)	→	from^#(0)	(20)
plus^#(s(X),Y)	→	plus^#(X,Y)	(21)
times^#(s(X),Y)	→	plus^#(Y,times(X,Y))	(22)
times^#(s(X),Y)	→	times^#(X,Y)	(23)
square^#(X)	→	times^#(X,X)	(24)

and the following rules have been deleted.

1.1.1.1 Pair and Rule Removal

Some pairs and rules have been removed and it remains to prove infiniteness of the remaing problem. The following pairs have been deleted. and the following rules have been deleted.

from(X)	→	cons(X,from(s(X)))	(1)
2ndspos(0,Z)	→	rnil	(2)
2ndspos(s(N),cons(X,Z))	→	2ndspos(s(N),cons2(X,Z))	(3)
2ndspos(s(N),cons2(X,cons(Y,Z)))	→	rcons(posrecip(Y),2ndsneg(N,Z))	(4)
2ndsneg(0,Z)	→	rnil	(5)
2ndsneg(s(N),cons(X,Z))	→	2ndsneg(s(N),cons2(X,Z))	(6)
2ndsneg(s(N),cons2(X,cons(Y,Z)))	→	rcons(negrecip(Y),2ndspos(N,Z))	(7)
pi(X)	→	2ndspos(X,from(0))	(8)
plus(0,Y)	→	Y	(9)
plus(s(X),Y)	→	s(plus(X,Y))	(10)
times(0,Y)	→	0	(11)
times(s(X),Y)	→	plus(Y,times(X,Y))	(12)
square(X)	→	times(X,X)	(13)

1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.

from^#(s(z0))

→

from^#(s(s(z0)))

(25)

1.1.1.1.1.1.1 Instantiation Processor

The pairs are instantiated to the following pairs.

from^#(s(s(z0)))

→

from^#(s(s(s(z0))))

(26)

1.1.1.1.1.1.1.1 Loop

The following loop proves infiniteness of the DP problem.

t₀	=	from^#(s(s(z0)))
	→_P	from^#(s(s(s(z0))))
	=	t₁

where t₁ = t₀σ and σ = {z0/s(z0)}