Certification Problem

Input (TPDB TRS_Standard/SK90/4.17)

The rewrite relation of the following TRS is considered.

fac(s(x)) → *(fac(p(s(x))),s(x)) (1)
p(s(0)) → 0 (2)
p(s(s(x))) → s(p(s(x))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[0] =
0 0 0
0 0 0
0 0 0
[fac(x1)] =
1 0 1
0 0 0
0 1 1
· x1 +
1 0 0
0 0 0
1 0 0
[*(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 1
1 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[p(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
fac(s(x)) → *(fac(p(s(x))),s(x)) (1)

1.1 Rule Removal

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

1.1.1 R is empty

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