YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Wait(Left(x0)) → Begin(x0)
a(x0) → b(x0)

Proof

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 0 · x1 + -∞
[Aa(x1)] = 15 · x1 + -∞
[Ab(x1)] = 0 · x1 + -∞
[Wait(x1)] = 7 · x1 + -∞
[Left(x1)] = 1 · x1 + -∞
[Begin(x1)] = 8 · x1 + -∞
[a(x1)] = 15 · x1 + -∞
the rules
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Wait(Left(x0)) → Begin(x0)
remain.

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
[b(x1)] =
1 2
0 3
· x1 +
-∞ -∞
-∞ -∞
[Aa(x1)] =
0 -∞
0 -∞
· x1 +
-∞ -∞
-∞ -∞
[Ab(x1)] =
1 0
2 3
· x1 +
-∞ -∞
-∞ -∞
[Wait(x1)] =
3 1
-∞ 3
· x1 +
-∞ -∞
-∞ -∞
[Left(x1)] =
0 0
0 3
· x1 +
-∞ -∞
-∞ -∞
[Begin(x1)] =
0 -∞
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 0
-∞ -∞
· x1 +
-∞ -∞
-∞ -∞
the rules
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(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
[b(x1)] =
0 1
-∞ 0
· x1 +
-∞ -∞
-∞ -∞
[Aa(x1)] =
0 1
-∞ 3
· x1 +
-∞ -∞
-∞ -∞
[Ab(x1)] =
0 -∞
1 -∞
· x1 +
-∞ -∞
-∞ -∞
[Left(x1)] =
0 1
1 2
· x1 +
-∞ -∞
-∞ -∞
[a(x1)] =
0 -∞
0 1
· x1 +
-∞ -∞
-∞ -∞
the rule
Ab(Left(x0)) → Left(b(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
[b(x1)] =
0 -∞
0 -∞
· x1 +
-∞ -∞
-∞ -∞
[Ab(x1)] =
1 0
0 0
· x1 +
-∞ -∞
-∞ -∞
[Left(x1)] =
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.