YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
Begin(b(b(x0))) |
→ |
Wait(Right1(x0)) |
|
Begin(b(x0)) |
→ |
Wait(Right2(x0)) |
|
Right1(b(End(x0))) |
→ |
Left(a(End(x0))) |
|
Right2(b(b(End(x0)))) |
→ |
Left(a(End(x0))) |
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
|
Wait(Left(x0)) |
→ |
Begin(x0) |
|
a(x0) |
→ |
b(b(x0)) |
|
b(b(b(x0))) |
→ |
a(x0) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [Right2(x1)] |
= |
5 ·
x1 +
-∞
|
| [Aa(x1)] |
= |
12 ·
x1 +
-∞
|
| [Begin(x1)] |
= |
6 ·
x1 +
-∞
|
| [Wait(x1)] |
= |
5 ·
x1 +
-∞
|
| [End(x1)] |
= |
11 ·
x1 +
-∞
|
| [b(x1)] |
= |
4 ·
x1 +
-∞
|
| [Left(x1)] |
= |
1 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
4 ·
x1 +
-∞
|
| [a(x1)] |
= |
12 ·
x1 +
-∞
|
| [Right1(x1)] |
= |
9 ·
x1 +
-∞
|
the
rules
|
Begin(b(b(x0))) |
→ |
Wait(Right1(x0)) |
|
Begin(b(x0)) |
→ |
Wait(Right2(x0)) |
|
Right1(b(End(x0))) |
→ |
Left(a(End(x0))) |
|
Right2(b(b(End(x0)))) |
→ |
Left(a(End(x0))) |
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
|
Wait(Left(x0)) |
→ |
Begin(x0) |
|
b(b(b(x0))) |
→ |
a(x0) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [Right2(x1)] |
= |
4 ·
x1 +
-∞
|
| [Aa(x1)] |
= |
3 ·
x1 +
-∞
|
| [Begin(x1)] |
= |
0 ·
x1 +
-∞
|
| [Wait(x1)] |
= |
1 ·
x1 +
-∞
|
| [End(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
9 ·
x1 +
-∞
|
| [Left(x1)] |
= |
1 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
9 ·
x1 +
-∞
|
| [a(x1)] |
= |
3 ·
x1 +
-∞
|
| [Right1(x1)] |
= |
3 ·
x1 +
-∞
|
the
rules
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
remain.
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(Ab) |
= |
1 |
|
weight(Ab) |
= |
1 |
|
|
|
| prec(Aa) |
= |
4 |
|
weight(Aa) |
= |
1 |
|
|
|
| prec(Left) |
= |
0 |
|
weight(Left) |
= |
1 |
|
|
|
| prec(a) |
= |
7 |
|
weight(a) |
= |
1 |
|
|
|
| prec(Right2) |
= |
6 |
|
weight(Right2) |
= |
1 |
|
|
|
| prec(Right1) |
= |
5 |
|
weight(Right1) |
= |
1 |
|
|
|
| prec(b) |
= |
0 |
|
weight(b) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.