YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(x0)) → b(x0)
a(b(x0)) → b(a(x0))
c(b(b(x0))) → a(c(x0))
b(b(x0)) → a(a(a(x0)))
a(c(x0)) → c(a(b(x0)))

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 3 · x1 + -∞
[a(x1)] = 2 · x1 + -∞
[c(x1)] = 10 · x1 + -∞
the rules
a(b(x0)) → b(a(x0))
b(b(x0)) → a(a(a(x0)))
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 8 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
the rule
a(b(x0)) → b(a(x0))
remains.

1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1 R is empty

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