Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-141)

The relative rewrite relation R/S is considered where R is the following TRS

a(c(c(x1)))	→	c(c(a(x1)))	(1)
a(b(b(x1)))	→	b(b(a(x1)))	(2)
a(a(c(x1)))	→	b(b(a(x1)))	(3)
b(c(a(x1)))	→	a(c(c(x1)))	(4)

and S is the following TRS.

b(a(a(x1)))	→	a(a(a(x1)))	(5)
a(c(a(x1)))	→	a(c(b(x1)))	(6)

Property / Task

Prove or disprove termination.

Answer / Result

No.

Proof (by AProVE @ termCOMP 2023)

1 Loop

The following loop proves that R/S is not relative terminating.

t₀	=	b(b(a(a(c(a(c(a(x1))))))))
	→_S	b(b(a(a(c(b(c(a(x1))))))))
	→_R	b(b(a(a(c(a(c(c(x1))))))))
	→_R	b(b(a(a(c(c(c(a(x1))))))))
	→_S	b(a(a(a(c(c(c(a(x1))))))))
	→_R	b(a(a(c(c(a(c(a(x1))))))))
	→_S	a(a(a(c(c(a(c(a(x1))))))))
	→_R	a(b(b(a(c(a(c(a(x1))))))))
	→_R	b(b(a(a(c(a(c(a(x1))))))))
	=	t₈

where t₈ = t₀σ and σ = {x1/x1}