YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
Begin(a(x0)) |
→ |
Wait(Right1(x0)) |
|
Begin(b(x0)) |
→ |
Wait(Right2(x0)) |
|
Begin(c(x0)) |
→ |
Wait(Right3(x0)) |
|
Begin(d(d(x0))) |
→ |
Wait(Right4(x0)) |
|
Begin(d(x0)) |
→ |
Wait(Right5(x0)) |
|
Right1(a(End(x0))) |
→ |
Left(b(c(End(x0)))) |
|
Right2(b(End(x0))) |
→ |
Left(c(d(End(x0)))) |
|
Right3(c(End(x0))) |
→ |
Left(d(d(d(End(x0))))) |
|
Right4(d(End(x0))) |
→ |
Left(a(c(End(x0)))) |
|
Right5(d(d(End(x0)))) |
→ |
Left(a(c(End(x0)))) |
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right3(a(x0)) |
→ |
Aa(Right3(x0)) |
|
Right4(a(x0)) |
→ |
Aa(Right4(x0)) |
|
Right5(a(x0)) |
→ |
Aa(Right5(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Right3(b(x0)) |
→ |
Ab(Right3(x0)) |
|
Right4(b(x0)) |
→ |
Ab(Right4(x0)) |
|
Right5(b(x0)) |
→ |
Ab(Right5(x0)) |
|
Right1(c(x0)) |
→ |
Ac(Right1(x0)) |
|
Right2(c(x0)) |
→ |
Ac(Right2(x0)) |
|
Right3(c(x0)) |
→ |
Ac(Right3(x0)) |
|
Right4(c(x0)) |
→ |
Ac(Right4(x0)) |
|
Right5(c(x0)) |
→ |
Ac(Right5(x0)) |
|
Right1(d(x0)) |
→ |
Ad(Right1(x0)) |
|
Right2(d(x0)) |
→ |
Ad(Right2(x0)) |
|
Right3(d(x0)) |
→ |
Ad(Right3(x0)) |
|
Right4(d(x0)) |
→ |
Ad(Right4(x0)) |
|
Right5(d(x0)) |
→ |
Ad(Right5(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
|
Ac(Left(x0)) |
→ |
Left(c(x0)) |
|
Ad(Left(x0)) |
→ |
Left(d(x0)) |
|
Wait(Left(x0)) |
→ |
Begin(x0) |
|
a(a(x0)) |
→ |
b(c(x0)) |
|
b(b(x0)) |
→ |
c(d(x0)) |
|
c(c(x0)) |
→ |
d(d(d(x0))) |
|
d(d(d(x0))) |
→ |
a(c(x0)) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [b(x1)] |
= |
10 ·
x1 +
-∞
|
| [Ac(x1)] |
= |
12 ·
x1 +
-∞
|
| [d(x1)] |
= |
8 ·
x1 +
-∞
|
| [Begin(x1)] |
= |
8 ·
x1 +
-∞
|
| [Wait(x1)] |
= |
8 ·
x1 +
-∞
|
| [Ad(x1)] |
= |
8 ·
x1 +
-∞
|
| [Right2(x1)] |
= |
10 ·
x1 +
-∞
|
| [Right5(x1)] |
= |
7 ·
x1 +
-∞
|
| [a(x1)] |
= |
11 ·
x1 +
-∞
|
| [Right3(x1)] |
= |
12 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
10 ·
x1 +
-∞
|
| [Left(x1)] |
= |
0 ·
x1 +
-∞
|
| [End(x1)] |
= |
6 ·
x1 +
-∞
|
| [Right4(x1)] |
= |
15 ·
x1 +
-∞
|
| [Aa(x1)] |
= |
11 ·
x1 +
-∞
|
| [c(x1)] |
= |
12 ·
x1 +
-∞
|
| [Right1(x1)] |
= |
11 ·
x1 +
-∞
|
the
rules
|
Begin(a(x0)) |
→ |
Wait(Right1(x0)) |
|
Begin(b(x0)) |
→ |
Wait(Right2(x0)) |
|
Begin(c(x0)) |
→ |
Wait(Right3(x0)) |
|
Right1(a(End(x0))) |
→ |
Left(b(c(End(x0)))) |
|
Right2(b(End(x0))) |
→ |
Left(c(d(End(x0)))) |
|
Right3(c(End(x0))) |
→ |
Left(d(d(d(End(x0))))) |
|
Right4(d(End(x0))) |
→ |
Left(a(c(End(x0)))) |
|
Right5(d(d(End(x0)))) |
→ |
Left(a(c(End(x0)))) |
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right3(a(x0)) |
→ |
Aa(Right3(x0)) |
|
Right4(a(x0)) |
→ |
Aa(Right4(x0)) |
|
Right5(a(x0)) |
→ |
Aa(Right5(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Right3(b(x0)) |
→ |
Ab(Right3(x0)) |
|
Right4(b(x0)) |
→ |
Ab(Right4(x0)) |
|
Right5(b(x0)) |
→ |
Ab(Right5(x0)) |
|
Right1(c(x0)) |
→ |
Ac(Right1(x0)) |
|
Right2(c(x0)) |
→ |
Ac(Right2(x0)) |
|
Right3(c(x0)) |
→ |
Ac(Right3(x0)) |
|
Right4(c(x0)) |
→ |
Ac(Right4(x0)) |
|
Right5(c(x0)) |
→ |
Ac(Right5(x0)) |
|
Right1(d(x0)) |
→ |
Ad(Right1(x0)) |
|
Right2(d(x0)) |
→ |
Ad(Right2(x0)) |
|
Right3(d(x0)) |
→ |
Ad(Right3(x0)) |
|
Right4(d(x0)) |
→ |
Ad(Right4(x0)) |
|
Right5(d(x0)) |
→ |
Ad(Right5(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
|
Ac(Left(x0)) |
→ |
Left(c(x0)) |
|
Ad(Left(x0)) |
→ |
Left(d(x0)) |
|
Wait(Left(x0)) |
→ |
Begin(x0) |
|
a(a(x0)) |
→ |
b(c(x0)) |
|
b(b(x0)) |
→ |
c(d(x0)) |
|
c(c(x0)) |
→ |
d(d(d(x0))) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [b(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ac(x1)] |
= |
0 ·
x1 +
-∞
|
| [d(x1)] |
= |
0 ·
x1 +
-∞
|
| [Begin(x1)] |
= |
0 ·
x1 +
-∞
|
| [Wait(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ad(x1)] |
= |
0 ·
x1 +
-∞
|
| [Right2(x1)] |
= |
0 ·
x1 +
-∞
|
| [Right5(x1)] |
= |
1 ·
x1 +
-∞
|
| [a(x1)] |
= |
1 ·
x1 +
-∞
|
| [Right3(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
0 ·
x1 +
-∞
|
| [Left(x1)] |
= |
0 ·
x1 +
-∞
|
| [End(x1)] |
= |
0 ·
x1 +
-∞
|
| [Right4(x1)] |
= |
1 ·
x1 +
-∞
|
| [Aa(x1)] |
= |
1 ·
x1 +
-∞
|
| [c(x1)] |
= |
0 ·
x1 +
-∞
|
| [Right1(x1)] |
= |
1 ·
x1 +
-∞
|
the
rules
|
Begin(a(x0)) |
→ |
Wait(Right1(x0)) |
|
Begin(b(x0)) |
→ |
Wait(Right2(x0)) |
|
Begin(c(x0)) |
→ |
Wait(Right3(x0)) |
|
Right2(b(End(x0))) |
→ |
Left(c(d(End(x0)))) |
|
Right3(c(End(x0))) |
→ |
Left(d(d(d(End(x0))))) |
|
Right4(d(End(x0))) |
→ |
Left(a(c(End(x0)))) |
|
Right5(d(d(End(x0)))) |
→ |
Left(a(c(End(x0)))) |
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right3(a(x0)) |
→ |
Aa(Right3(x0)) |
|
Right4(a(x0)) |
→ |
Aa(Right4(x0)) |
|
Right5(a(x0)) |
→ |
Aa(Right5(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Right3(b(x0)) |
→ |
Ab(Right3(x0)) |
|
Right4(b(x0)) |
→ |
Ab(Right4(x0)) |
|
Right5(b(x0)) |
→ |
Ab(Right5(x0)) |
|
Right1(c(x0)) |
→ |
Ac(Right1(x0)) |
|
Right2(c(x0)) |
→ |
Ac(Right2(x0)) |
|
Right3(c(x0)) |
→ |
Ac(Right3(x0)) |
|
Right4(c(x0)) |
→ |
Ac(Right4(x0)) |
|
Right5(c(x0)) |
→ |
Ac(Right5(x0)) |
|
Right1(d(x0)) |
→ |
Ad(Right1(x0)) |
|
Right2(d(x0)) |
→ |
Ad(Right2(x0)) |
|
Right3(d(x0)) |
→ |
Ad(Right3(x0)) |
|
Right4(d(x0)) |
→ |
Ad(Right4(x0)) |
|
Right5(d(x0)) |
→ |
Ad(Right5(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
|
Ac(Left(x0)) |
→ |
Left(c(x0)) |
|
Ad(Left(x0)) |
→ |
Left(d(x0)) |
|
Wait(Left(x0)) |
→ |
Begin(x0) |
|
b(b(x0)) |
→ |
c(d(x0)) |
|
c(c(x0)) |
→ |
d(d(d(x0))) |
remain.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [b(x1)] |
= |
4 ·
x1 +
-∞
|
| [Ac(x1)] |
= |
1 ·
x1 +
-∞
|
| [d(x1)] |
= |
0 ·
x1 +
-∞
|
| [Begin(x1)] |
= |
0 ·
x1 +
-∞
|
| [Wait(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ad(x1)] |
= |
0 ·
x1 +
-∞
|
| [Right2(x1)] |
= |
3 ·
x1 +
-∞
|
| [Right5(x1)] |
= |
4 ·
x1 +
-∞
|
| [a(x1)] |
= |
2 ·
x1 +
-∞
|
| [Right3(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
4 ·
x1 +
-∞
|
| [Left(x1)] |
= |
0 ·
x1 +
-∞
|
| [End(x1)] |
= |
0 ·
x1 +
-∞
|
| [Right4(x1)] |
= |
4 ·
x1 +
-∞
|
| [Aa(x1)] |
= |
2 ·
x1 +
-∞
|
| [c(x1)] |
= |
1 ·
x1 +
-∞
|
| [Right1(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
Right1(a(x0)) |
→ |
Aa(Right1(x0)) |
|
Right2(a(x0)) |
→ |
Aa(Right2(x0)) |
|
Right3(a(x0)) |
→ |
Aa(Right3(x0)) |
|
Right4(a(x0)) |
→ |
Aa(Right4(x0)) |
|
Right5(a(x0)) |
→ |
Aa(Right5(x0)) |
|
Right1(b(x0)) |
→ |
Ab(Right1(x0)) |
|
Right2(b(x0)) |
→ |
Ab(Right2(x0)) |
|
Right3(b(x0)) |
→ |
Ab(Right3(x0)) |
|
Right4(b(x0)) |
→ |
Ab(Right4(x0)) |
|
Right5(b(x0)) |
→ |
Ab(Right5(x0)) |
|
Right1(c(x0)) |
→ |
Ac(Right1(x0)) |
|
Right2(c(x0)) |
→ |
Ac(Right2(x0)) |
|
Right3(c(x0)) |
→ |
Ac(Right3(x0)) |
|
Right4(c(x0)) |
→ |
Ac(Right4(x0)) |
|
Right5(c(x0)) |
→ |
Ac(Right5(x0)) |
|
Right1(d(x0)) |
→ |
Ad(Right1(x0)) |
|
Right2(d(x0)) |
→ |
Ad(Right2(x0)) |
|
Right3(d(x0)) |
→ |
Ad(Right3(x0)) |
|
Right4(d(x0)) |
→ |
Ad(Right4(x0)) |
|
Right5(d(x0)) |
→ |
Ad(Right5(x0)) |
|
Aa(Left(x0)) |
→ |
Left(a(x0)) |
|
Ab(Left(x0)) |
→ |
Left(b(x0)) |
|
Ac(Left(x0)) |
→ |
Left(c(x0)) |
|
Ad(Left(x0)) |
→ |
Left(d(x0)) |
|
Wait(Left(x0)) |
→ |
Begin(x0) |
remain.
1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(Ad) |
= |
2 |
|
weight(Ad) |
= |
1 |
|
|
|
| prec(Ac) |
= |
1 |
|
weight(Ac) |
= |
1 |
|
|
|
| prec(Ab) |
= |
1 |
|
weight(Ab) |
= |
1 |
|
|
|
| prec(Aa) |
= |
1 |
|
weight(Aa) |
= |
1 |
|
|
|
| prec(Left) |
= |
0 |
|
weight(Left) |
= |
1 |
|
|
|
| prec(Right5) |
= |
3 |
|
weight(Right5) |
= |
1 |
|
|
|
| prec(Right4) |
= |
7 |
|
weight(Right4) |
= |
1 |
|
|
|
| prec(d) |
= |
0 |
|
weight(d) |
= |
1 |
|
|
|
| prec(Right3) |
= |
7 |
|
weight(Right3) |
= |
1 |
|
|
|
| prec(c) |
= |
0 |
|
weight(c) |
= |
1 |
|
|
|
| prec(Right2) |
= |
8 |
|
weight(Right2) |
= |
1 |
|
|
|
| prec(b) |
= |
0 |
|
weight(b) |
= |
1 |
|
|
|
| prec(Wait) |
= |
9 |
|
weight(Wait) |
= |
0 |
|
|
|
| prec(Right1) |
= |
8 |
|
weight(Right1) |
= |
1 |
|
|
|
| prec(Begin) |
= |
0 |
|
weight(Begin) |
= |
1 |
|
|
|
| prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.