YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
R(E(x0)) |
→ |
L(E(x0)) |
|
a(L(x0)) |
→ |
L(Aa(x0)) |
|
b(L(x0)) |
→ |
L(Ab(x0)) |
|
R(Aa(x0)) |
→ |
a(R(x0)) |
|
R(Ab(x0)) |
→ |
b(R(x0)) |
|
a(L(x0)) |
→ |
b(R(x0)) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [Aa(x1)] |
= |
13 ·
x1 +
-∞
|
| [R(x1)] |
= |
14 ·
x1 +
-∞
|
| [a(x1)] |
= |
13 ·
x1 +
-∞
|
| [b(x1)] |
= |
10 ·
x1 +
-∞
|
| [E(x1)] |
= |
8 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
10 ·
x1 +
-∞
|
| [L(x1)] |
= |
14 ·
x1 +
-∞
|
the
rules
|
R(E(x0)) |
→ |
L(E(x0)) |
|
a(L(x0)) |
→ |
L(Aa(x0)) |
|
b(L(x0)) |
→ |
L(Ab(x0)) |
|
R(Aa(x0)) |
→ |
a(R(x0)) |
|
R(Ab(x0)) |
→ |
b(R(x0)) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [Aa(x1)] |
= |
0 ·
x1 +
-∞
|
| [R(x1)] |
= |
1 ·
x1 +
-∞
|
| [a(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
0 ·
x1 +
-∞
|
| [E(x1)] |
= |
0 ·
x1 +
-∞
|
| [Ab(x1)] |
= |
0 ·
x1 +
-∞
|
| [L(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
a(L(x0)) |
→ |
L(Aa(x0)) |
|
b(L(x0)) |
→ |
L(Ab(x0)) |
|
R(Aa(x0)) |
→ |
a(R(x0)) |
|
R(Ab(x0)) |
→ |
b(R(x0)) |
remain.
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(Ab) |
= |
0 |
|
weight(Ab) |
= |
1 |
|
|
|
| prec(b) |
= |
1 |
|
weight(b) |
= |
1 |
|
|
|
| prec(Aa) |
= |
3 |
|
weight(Aa) |
= |
1 |
|
|
|
| prec(a) |
= |
1 |
|
weight(a) |
= |
1 |
|
|
|
| prec(L) |
= |
0 |
|
weight(L) |
= |
1 |
|
|
|
| prec(R) |
= |
2 |
|
weight(R) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.