YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Rule Removal

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

1.1.1 Rule Removal

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

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the arctic semiring over the integers
[a(x1)] =
1 2
0 0
· x1 +
-∞ -∞
-∞ -∞
[c(x1)] =
0 1
-∞ 0
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
0 -∞
1 3
· x1 +
-∞ -∞
-∞ -∞
all rules could be removed.

1.1.1.1.1 R is empty

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