Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z071)

The rewrite relation of the following TRS is considered.

R(x1)	→	r(x1)	(1)
r(p(x1))	→	p(p(r(P(x1))))	(2)
r(r(x1))	→	x1	(3)
r(P(P(x1)))	→	P(P(r(x1)))	(4)
p(P(x1))	→	x1	(5)
P(p(x1))	→	x1	(6)
r(R(x1))	→	x1	(7)
R(r(x1))	→	x1	(8)

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.

r^#(p(x1))	→	p^#(r(P(x1)))	(9)
r^#(P(P(x1)))	→	P^#(P(r(x1)))	(10)
r^#(P(P(x1)))	→	P^#(r(x1))	(11)
R^#(x1)	→	r^#(x1)	(12)
r^#(p(x1))	→	r^#(P(x1))	(13)
r^#(p(x1))	→	p^#(p(r(P(x1))))	(14)
r^#(P(P(x1)))	→	r^#(x1)	(15)
r^#(p(x1))	→	P^#(x1)	(16)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

r^#(P(P(x1))) → r^#(x1) (15)

r^#(p(x1)) → r^#(P(x1)) (13)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[P^#(x₁)] = 0

[r(x₁)] = 0

[p^#(x₁)] = 0

[R^#(x₁)] = 0

[p(x₁)] = x₁ + 2

[R(x₁)] = 0

[r^#(x₁)] = x₁ + 0

[P(x₁)] = x₁ + 1

together with the usable rule
P(p(x1)) → x1 (6)
(w.r.t. the implicit argument filter of the reduction pair), the pairs

r^#(P(P(x1))) → r^#(x1) (15)

r^#(p(x1)) → r^#(P(x1)) (13)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.