YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
r1(a(x0)) |
→ |
a(a(a(r1(x0)))) |
|
r2(a(x0)) |
→ |
a(a(a(r2(x0)))) |
|
a(l1(x0)) |
→ |
l1(a(a(a(x0)))) |
|
a(a(l2(x0))) |
→ |
l2(a(a(x0))) |
|
r1(b(x0)) |
→ |
l1(b(x0)) |
|
r2(b(x0)) |
→ |
l2(a(b(x0))) |
|
b(l1(x0)) |
→ |
b(r2(x0)) |
|
b(l2(x0)) |
→ |
b(r1(x0)) |
|
a(a(x0)) |
→ |
x0 |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [l2(x1)] |
= |
·
x1 +
|
| [r1(x1)] |
= |
·
x1 +
|
| [l1(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [a(x1)] |
= |
·
x1 +
|
| [r2(x1)] |
= |
·
x1 +
|
the
rules
|
r1(a(x0)) |
→ |
a(a(a(r1(x0)))) |
|
r2(a(x0)) |
→ |
a(a(a(r2(x0)))) |
|
a(l1(x0)) |
→ |
l1(a(a(a(x0)))) |
|
a(a(l2(x0))) |
→ |
l2(a(a(x0))) |
|
r1(b(x0)) |
→ |
l1(b(x0)) |
|
r2(b(x0)) |
→ |
l2(a(b(x0))) |
|
b(l1(x0)) |
→ |
b(r2(x0)) |
|
a(a(x0)) |
→ |
x0 |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [l2(x1)] |
= |
0 ·
x1 +
-∞
|
| [r1(x1)] |
= |
14 ·
x1 +
-∞
|
| [l1(x1)] |
= |
14 ·
x1 +
-∞
|
| [b(x1)] |
= |
8 ·
x1 +
-∞
|
| [a(x1)] |
= |
0 ·
x1 +
-∞
|
| [r2(x1)] |
= |
14 ·
x1 +
-∞
|
the
rules
|
r1(a(x0)) |
→ |
a(a(a(r1(x0)))) |
|
r2(a(x0)) |
→ |
a(a(a(r2(x0)))) |
|
a(l1(x0)) |
→ |
l1(a(a(a(x0)))) |
|
a(a(l2(x0))) |
→ |
l2(a(a(x0))) |
|
r1(b(x0)) |
→ |
l1(b(x0)) |
|
b(l1(x0)) |
→ |
b(r2(x0)) |
|
a(a(x0)) |
→ |
x0 |
remain.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [l2(x1)] |
= |
0 ·
x1 +
-∞
|
| [r1(x1)] |
= |
6 ·
x1 +
-∞
|
| [l1(x1)] |
= |
4 ·
x1 +
-∞
|
| [b(x1)] |
= |
12 ·
x1 +
-∞
|
| [a(x1)] |
= |
0 ·
x1 +
-∞
|
| [r2(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
r1(a(x0)) |
→ |
a(a(a(r1(x0)))) |
|
r2(a(x0)) |
→ |
a(a(a(r2(x0)))) |
|
a(l1(x0)) |
→ |
l1(a(a(a(x0)))) |
|
a(a(l2(x0))) |
→ |
l2(a(a(x0))) |
|
a(a(x0)) |
→ |
x0 |
remain.
1.1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(r1(x0)) |
→ |
r1(a(a(a(x0)))) |
|
a(r2(x0)) |
→ |
r2(a(a(a(x0)))) |
|
l1(a(x0)) |
→ |
a(a(a(l1(x0)))) |
|
l2(a(a(x0))) |
→ |
a(a(l2(x0))) |
|
a(a(x0)) |
→ |
x0 |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [l2(x1)] |
= |
·
x1 +
|
| [r1(x1)] |
= |
·
x1 +
|
| [l1(x1)] |
= |
·
x1 +
|
| [a(x1)] |
= |
·
x1 +
|
| [r2(x1)] |
= |
·
x1 +
|
the
rules
|
a(r2(x0)) |
→ |
r2(a(a(a(x0)))) |
|
l1(a(x0)) |
→ |
a(a(a(l1(x0)))) |
|
l2(a(a(x0))) |
→ |
a(a(l2(x0))) |
|
a(a(x0)) |
→ |
x0 |
remain.
1.1.1.1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
r2(a(x0)) |
→ |
a(a(a(r2(x0)))) |
|
a(l1(x0)) |
→ |
l1(a(a(a(x0)))) |
|
a(a(l2(x0))) |
→ |
l2(a(a(x0))) |
|
a(a(x0)) |
→ |
x0 |
1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [l2(x1)] |
= |
·
x1 +
|
| [l1(x1)] |
= |
·
x1 +
|
| [a(x1)] |
= |
·
x1 +
|
| [r2(x1)] |
= |
·
x1 +
|
the
rules
|
a(l1(x0)) |
→ |
l1(a(a(a(x0)))) |
|
a(a(l2(x0))) |
→ |
l2(a(a(x0))) |
|
a(a(x0)) |
→ |
x0 |
remain.
1.1.1.1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(l2) |
= |
0 |
|
weight(l2) |
= |
1 |
|
|
|
| prec(l1) |
= |
0 |
|
weight(l1) |
= |
1 |
|
|
|
| prec(a) |
= |
1 |
|
weight(a) |
= |
0 |
|
|
|
all rules could be removed.
1.1.1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.