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 ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] = 0 · x1 + -∞
[q(x1)] = 0 · x1 + -∞
[P(x1)] = 13 · x1 + -∞
[Q(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
P(x1) → Q(Q(p(x1))) (1)
p(P(x1)) → x1 (8)
P(p(x1)) → x1 (9)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] = 8 · x1 + -∞
[q(x1)] = 1 · x1 + -∞
[Q(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
p(p(x1)) → q(q(x1)) (2)
q(Q(x1)) → x1 (6)
Q(q(x1)) → x1 (7)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[p(x1)] =
1 1 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[q(x1)] =
1 0 0
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[Q(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
p(Q(Q(x1))) → Q(Q(p(x1))) (3)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(q) = 2 weight(q) = 2
prec(Q) = 3 weight(Q) = 2
prec(p) = 0 weight(p) = 2
all of the following rules can be deleted.
Q(p(q(x1))) → q(p(Q(x1))) (4)
q(q(p(x1))) → p(q(q(x1))) (5)

1.1.1.1.1 R is empty

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