Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z103)

The rewrite relation of the following TRS is considered.

a(d(x1)) → d(b(x1)) (1)
a(x1) → b(b(b(x1))) (2)
b(d(b(x1))) → a(c(x1)) (3)
c(x1) → d(x1) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[a(x1)] =
1 0 0 0 0
0 0 0 0 0
0 1 1 0 0
0 0 0 1 0
0 0 0 0 1
· x1 +
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[c(x1)] =
1 1 1 1 1
0 1 0 0 0
0 0 1 1 1
0 1 1 1 1
0 1 1 1 1
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[d(x1)] =
1 1 1 1 1
0 1 0 0 0
0 0 1 1 1
0 1 1 1 1
0 1 1 1 1
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[b(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 1
0 1 1 0 0
0 0 0 1 0
· x1 +
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
a(x1) → b(b(b(x1))) (2)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) = 2 weight(c) = 5
prec(b) = 3 weight(b) = 4
prec(a) = 1 weight(a) = 4
prec(d) = 0 weight(d) = 1
all of the following rules can be deleted.
a(d(x1)) → d(b(x1)) (1)
b(d(b(x1))) → a(c(x1)) (3)
c(x1) → d(x1) (4)

1.1.1 R is empty

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