Certification Problem

Input (TPDB TRS_Standard/SK90/2.15)

The rewrite relation of the following TRS is considered.

f(0) → 1 (1)
f(s(x)) → g(f(x)) (2)
g(x) → +(x,s(x)) (3)
f(s(x)) → +(f(x),s(f(x))) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[g(x1)] =
3 4
0 1
· x1 +
5 0
0 0
[f(x1)] =
1 1
0 0
· x1 +
0 0
0 0
[s(x1)] =
1 1
2 2
· x1 +
2 0
3 0
[+(x1, x2)] =
1 1
0 0
· x1 +
2 0
0 0
· x2 +
0 0
0 0
[0] =
0 0
2 0
[1] =
2 0
0 0
all of the following rules can be deleted.
g(x) → +(x,s(x)) (3)
f(s(x)) → +(f(x),s(f(x))) (4)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 2 weight(g) = 1
prec(s) = 0 weight(s) = 1
prec(1) = 6 weight(1) = 1
prec(f) = 7 weight(f) = 0
prec(0) = 4 weight(0) = 1
all of the following rules can be deleted.
f(0) → 1 (1)
f(s(x)) → g(f(x)) (2)

1.1.1 R is empty

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