YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(b(b(a(x0)))) b(a(a(b(x0))))
b(a(b(x0))) a(b(b(b(x0))))

Proof

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 0
0 0 1
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
2 0 1
0 0 0
0 3 0
· x1 +
1 0 0
0 0 0
0 0 0
the rule
b(a(b(x0))) a(b(b(b(x0))))
remains.

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight function
prec(b) = 1 weight(b) = 0
prec(a) = 0 weight(a) = 1
all rules could be removed.

1.1.1 R is empty

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