Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z125)

The rewrite relation of the following TRS is considered.

f(x1) → n(c(n(a(x1)))) (1)
c(f(x1)) → f(n(a(c(x1)))) (2)
n(a(x1)) → c(x1) (3)
c(c(x1)) → c(x1) (4)
n(s(x1)) → f(s(s(x1))) (5)
n(f(x1)) → f(n(x1)) (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
[s(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[a(x1)] =
1 0 0
0 0 0
0 0 0
· 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
[f(x1)] =
1 0 0
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[n(x1)] =
1 1 1
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
n(s(x1)) → f(s(s(x1))) (5)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[a(x1)] = 0 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[f(x1)] = 2 · x1 + -∞
[n(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
f(x1) → n(c(n(a(x1)))) (1)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
1 0 0
[n(x1)] =
1 0 1
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
n(f(x1)) → f(n(x1)) (6)

1.1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 1
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 1 1
1 1 1
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
[f(x1)] =
1 1 1
1 1 0
0 1 1
· x1 +
0 0 0
1 0 0
1 0 0
[n(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
c(f(x1)) → f(n(a(c(x1)))) (2)

1.1.1.1.1.1.1 R is empty

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