YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(a(b(x0))) → b(a(b(x0)))
b(a(x0)) → a(b(b(x0)))
b(c(a(x0))) → c(c(a(a(b(x0)))))

Proof

1 Rule Removal

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

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[a(x1)] = 1 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
the rule
b(a(x0)) → a(b(b(x0)))
remains.

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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