Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z075)

The rewrite relation of the following TRS is considered.

t(f(x1)) → t(c(n(x1))) (1)
n(f(x1)) → f(n(x1)) (2)
o(f(x1)) → f(o(x1)) (3)
n(s(x1)) → f(s(x1)) (4)
o(s(x1)) → f(s(x1)) (5)
c(f(x1)) → f(c(x1)) (6)
c(n(x1)) → n(c(x1)) (7)
c(o(x1)) → o(c(x1)) (8)
c(o(x1)) → o(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
[o(x1)] = 8 · x1 + -∞
[t(x1)] = 0 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[s(x1)] = 0 · x1 + -∞
[f(x1)] = 0 · x1 + -∞
[n(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
o(s(x1)) → f(s(x1)) (5)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(o) = 1 weight(o) = 2
prec(c) = 3 weight(c) = 2
prec(n) = 2 weight(n) = 1
prec(t) = 7 weight(t) = 2
prec(f) = 0 weight(f) = 6
all of the following rules can be deleted.
t(f(x1)) → t(c(n(x1))) (1)
n(f(x1)) → f(n(x1)) (2)
o(f(x1)) → f(o(x1)) (3)
c(f(x1)) → f(c(x1)) (6)
c(n(x1)) → n(c(x1)) (7)
c(o(x1)) → o(c(x1)) (8)
c(o(x1)) → o(x1) (9)

1.1.1.1 R is empty

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