Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup08)

The rewrite relation of the following TRS is considered.

a(a(a(a(b(b(x1)))))) b(b(a(a(b(b(x1)))))) (1)
b(b(a(a(x1)))) a(a(b(b(b(b(x1)))))) (2)
b(b(c(c(a(a(x1)))))) c(c(c(c(a(a(a(a(b(b(x1)))))))))) (3)

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 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[b(x1)] =
1 0 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 1
0 1 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[c(x1)] =
1 0 0 0 0
0 0 0 0 1
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
· 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
all of the following rules can be deleted.
b(b(c(c(a(a(x1)))))) c(c(c(c(a(a(a(a(b(b(x1)))))))))) (3)

1.1 Rule Removal

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

1.1.1 R is empty

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