Certification Problem

Input (TPDB TRS_Standard/AG01/#3.41)

The rewrite relation of the following TRS is considered.

p(s(x)) → x (1)
fac(0) → s(0) (2)
fac(s(x)) → times(s(x),fac(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
[times(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 0
0 1 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[p(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[fac(x1)] =
1 0 1
0 0 0
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[s(x1)] =
1 0 0
0 1 1
1 1 1
· x1 +
0 0 0
0 0 0
1 0 0
[0] =
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
fac(0) → s(0) (2)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(p) = 0 status(p) = [1] list-extension(p) = Lex
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
and the following Max-polynomial interpretation
[p(x1)] = max(0, 0 + 1 · x1)
[s(x1)] = max(0, 1 + 1 · x1)
all of the following rules can be deleted.
p(s(x)) → x (1)

1.1.1.1 R is empty

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