Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z067)

The rewrite relation of the following TRS is considered.

P(x1) → Q(Q(p(x1))) (1)
p(p(x1)) → q(q(x1)) (2)
p(Q(Q(x1))) → Q(Q(p(x1))) (3)
Q(p(q(x1))) → q(p(Q(x1))) (4)
q(q(p(x1))) → p(q(q(x1))) (5)
q(Q(x1)) → x1 (6)
Q(q(x1)) → x1 (7)
p(P(x1)) → x1 (8)
P(p(x1)) → x1 (9)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[Q(x1)] = x1 +
1
[P(x1)] = x1 +
4
[q(x1)] = x1 +
0
[p(x1)] = x1 +
1
all of the following rules can be deleted.
P(x1) → Q(Q(p(x1))) (1)
p(p(x1)) → q(q(x1)) (2)
q(Q(x1)) → x1 (6)
Q(q(x1)) → x1 (7)
p(P(x1)) → x1 (8)
P(p(x1)) → x1 (9)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
Q#(p(q(x1))) → Q#(x1) (10)
Q#(p(q(x1))) → q#(p(Q(x1))) (11)
Q#(p(q(x1))) → p#(Q(x1)) (12)
q#(q(p(x1))) → q#(x1) (13)
q#(q(p(x1))) → q#(q(x1)) (14)
q#(q(p(x1))) → p#(q(q(x1))) (15)
p#(Q(Q(x1))) → Q#(Q(p(x1))) (16)
p#(Q(Q(x1))) → Q#(p(x1)) (17)
p#(Q(Q(x1))) → p#(x1) (18)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[Q(x1)] = x1 +
1
[q(x1)] = x1 +
1
[p(x1)] = x1 +
1
[Q#(x1)] = x1 +
0
[q#(x1)] = x1 +
0
[p#(x1)] = x1 +
0
together with the usable rules
p(Q(Q(x1))) → Q(Q(p(x1))) (3)
Q(p(q(x1))) → q(p(Q(x1))) (4)
q(q(p(x1))) → p(q(q(x1))) (5)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
Q#(p(q(x1))) → Q#(x1) (10)
Q#(p(q(x1))) → p#(Q(x1)) (12)
q#(q(p(x1))) → q#(x1) (13)
q#(q(p(x1))) → q#(q(x1)) (14)
p#(Q(Q(x1))) → Q#(p(x1)) (17)
p#(Q(Q(x1))) → p#(x1) (18)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.