YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(a(x0)) → a(a(a(b(x0))))
a(b(x0)) → a(a(x0))
a(a(x0)) → b(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 0
0 1
· x1 +
0 0
1 0
[b(x1)] =
1 1
0 3
· x1 +
0 0
1 0
[c(x1)] =
1 0
0 0
· x1 +
0 0
0 0
the rules
a(b(x0)) → a(a(x0))
a(a(x0)) → b(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)] =
0 0
0 1
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
0 0
1 2
· x1 +
-∞ -∞
-∞ -∞
[c(x1)] =
0 0
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
the rule
a(a(x0)) → b(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)] =
2 0
1 0
· x1 +
-∞ -∞
-∞ -∞
[b(x1)] =
0 -∞
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[c(x1)] =
0 0
-∞ 0
· 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.