Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z119)

The rewrite relation of the following TRS is considered.

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

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(d) = 0 weight(d) = 2
prec(a) = 3 weight(a) = 2
prec(b) = 2 weight(b) = 2
all of the following rules can be deleted.
a(b(x1)) → b(d(x1)) (1)

1.1.1.1.1 R is empty

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