YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

a(x0) b(c(x0))
a(b(x0)) b(a(x0))
d(c(x0)) d(a(x0))
a(c(x0)) c(a(x0))

Proof

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[c(x1)] =
0 -∞
1 1
· x1 +
-∞ -∞
-∞ -∞
[d(x1)] =
0 0
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 -∞
0 0
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
0 -∞
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
the rules
a(x0) b(c(x0))
a(b(x0)) b(a(x0))
a(c(x0)) c(a(x0))
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[c(x1)] = 0 · x1 + -∞
[a(x1)] = 8 · x1 + -∞
[b(x1)] = 5 · x1 + -∞
the rules
a(b(x0)) b(a(x0))
a(c(x0)) c(a(x0))
remain.

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(c) = 0 weight(c) = 1
prec(a) = 1 weight(a) = 0
all rules could be removed.

1.1.1.1 R is empty

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