Certification Problem

Input (TPDB TRS_Standard/AG01/#3.36)

The rewrite relation of the following TRS is considered.

minus(x,0) → x (1)
minus(s(x),s(y)) → minus(x,y) (2)
f(0) → s(0) (3)
f(s(x)) → minus(s(x),g(f(x))) (4)
g(0) → 0 (5)
g(s(x)) → minus(s(x),f(g(x))) (6)

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
[g(x1)] =
1 0 1
0 1 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[minus(x1, x2)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
1 1 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 0
1 1 1
1 1 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
f(s(x)) → minus(s(x),g(f(x))) (4)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[g(x1)] =
1 1 1
1 1 1
1 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[minus(x1, x2)] =
1 0 0
1 1 1
0 0 1
· x1 +
1 0 1
0 0 1
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 1
1 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[0] =
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
1 0 1
1 1 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
g(0) → 0 (5)
g(s(x)) → minus(s(x),f(g(x))) (6)

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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