Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z121)

The rewrite relation of the following TRS is considered.

f(f(x1)) → b(b(b(x1))) (1)
a(f(x1)) → f(a(a(x1))) (2)
b(b(x1)) → c(c(a(c(x1)))) (3)
d(b(x1)) → d(a(b(x1))) (4)
c(c(x1)) → d(d(d(x1))) (5)
b(d(x1)) → d(b(x1)) (6)
c(d(d(x1))) → f(x1) (7)

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
[d(x1)] = 4 · x1 + -∞
[b(x1)] = 9 · x1 + -∞
[c(x1)] = 6 · x1 + -∞
[f(x1)] = 14 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
f(f(x1)) → b(b(b(x1))) (1)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[d(x1)] = 1 · x1 + -∞
[b(x1)] = 6 · x1 + -∞
[c(x1)] = 2 · x1 + -∞
[f(x1)] = 0 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
b(b(x1)) → c(c(a(c(x1)))) (3)
c(c(x1)) → d(d(d(x1))) (5)
c(d(d(x1))) → f(x1) (7)

1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(a) = 1 weight(a) = 0
prec(f) = 0 weight(f) = 2
all of the following rules can be deleted.
a(f(x1)) → f(a(a(x1))) (2)

1.1.1.1.1.1 R is empty

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