YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
r0(0(x0)) |
→ |
0(r0(x0)) |
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r0(m(x0)) |
→ |
m(r0(x0)) |
|
r1(0(x0)) |
→ |
0(r1(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r0(b(x0)) |
→ |
qr(0(b(x0))) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
m(qr(x0)) |
→ |
ql(m(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
|
b(ql(0(x0))) |
→ |
0(b(r0(x0))) |
|
b(ql(1(x0))) |
→ |
1(b(r1(x0))) |
Proof
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
0(r0(x0)) |
→ |
r0(0(x0)) |
|
1(r0(x0)) |
→ |
r0(1(x0)) |
|
m(r0(x0)) |
→ |
r0(m(x0)) |
|
0(r1(x0)) |
→ |
r1(0(x0)) |
|
1(r1(x0)) |
→ |
r1(1(x0)) |
|
m(r1(x0)) |
→ |
r1(m(x0)) |
|
b(r0(x0)) |
→ |
b(0(qr(x0))) |
|
b(r1(x0)) |
→ |
b(1(qr(x0))) |
|
qr(0(x0)) |
→ |
0(qr(x0)) |
|
qr(1(x0)) |
→ |
1(qr(x0)) |
|
qr(m(x0)) |
→ |
m(ql(x0)) |
|
ql(0(x0)) |
→ |
0(ql(x0)) |
|
ql(1(x0)) |
→ |
1(ql(x0)) |
|
0(ql(b(x0))) |
→ |
r0(b(0(x0))) |
|
1(ql(b(x0))) |
→ |
r1(b(1(x0))) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [r1(x1)] |
= |
·
x1 +
|
| [r0(x1)] |
= |
·
x1 +
|
| [m(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [0(x1)] |
= |
·
x1 +
|
| [ql(x1)] |
= |
·
x1 +
|
| [qr(x1)] |
= |
·
x1 +
|
| [1(x1)] |
= |
·
x1 +
|
the
rules
|
0(r0(x0)) |
→ |
r0(0(x0)) |
|
1(r0(x0)) |
→ |
r0(1(x0)) |
|
0(r1(x0)) |
→ |
r1(0(x0)) |
|
1(r1(x0)) |
→ |
r1(1(x0)) |
|
m(r1(x0)) |
→ |
r1(m(x0)) |
|
b(r0(x0)) |
→ |
b(0(qr(x0))) |
|
b(r1(x0)) |
→ |
b(1(qr(x0))) |
|
qr(0(x0)) |
→ |
0(qr(x0)) |
|
qr(1(x0)) |
→ |
1(qr(x0)) |
|
qr(m(x0)) |
→ |
m(ql(x0)) |
|
ql(0(x0)) |
→ |
0(ql(x0)) |
|
ql(1(x0)) |
→ |
1(ql(x0)) |
|
0(ql(b(x0))) |
→ |
r0(b(0(x0))) |
|
1(ql(b(x0))) |
→ |
r1(b(1(x0))) |
remain.
1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
r0(0(x0)) |
→ |
0(r0(x0)) |
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(0(x0)) |
→ |
0(r1(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r0(b(x0)) |
→ |
qr(0(b(x0))) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
m(qr(x0)) |
→ |
ql(m(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
|
b(ql(0(x0))) |
→ |
0(b(r0(x0))) |
|
b(ql(1(x0))) |
→ |
1(b(r1(x0))) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [r1(x1)] |
= |
·
x1 +
|
| [r0(x1)] |
= |
·
x1 +
|
| [m(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [0(x1)] |
= |
·
x1 +
|
| [ql(x1)] |
= |
·
x1 +
|
| [qr(x1)] |
= |
·
x1 +
|
| [1(x1)] |
= |
·
x1 +
|
the
rules
|
r0(0(x0)) |
→ |
0(r0(x0)) |
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(0(x0)) |
→ |
0(r1(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
m(qr(x0)) |
→ |
ql(m(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
|
b(ql(0(x0))) |
→ |
0(b(r0(x0))) |
|
b(ql(1(x0))) |
→ |
1(b(r1(x0))) |
remain.
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [r1(x1)] |
= |
1 ·
x1 +
-∞
|
| [r0(x1)] |
= |
0 ·
x1 +
-∞
|
| [m(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
0 ·
x1 +
-∞
|
| [0(x1)] |
= |
0 ·
x1 +
-∞
|
| [ql(x1)] |
= |
1 ·
x1 +
-∞
|
| [qr(x1)] |
= |
1 ·
x1 +
-∞
|
| [1(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
r0(0(x0)) |
→ |
0(r0(x0)) |
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(0(x0)) |
→ |
0(r1(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
m(qr(x0)) |
→ |
ql(m(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
|
b(ql(1(x0))) |
→ |
1(b(r1(x0))) |
remain.
1.1.1.1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
0(r0(x0)) |
→ |
r0(0(x0)) |
|
1(r0(x0)) |
→ |
r0(1(x0)) |
|
0(r1(x0)) |
→ |
r1(0(x0)) |
|
1(r1(x0)) |
→ |
r1(1(x0)) |
|
m(r1(x0)) |
→ |
r1(m(x0)) |
|
b(r1(x0)) |
→ |
b(1(qr(x0))) |
|
qr(0(x0)) |
→ |
0(qr(x0)) |
|
qr(1(x0)) |
→ |
1(qr(x0)) |
|
qr(m(x0)) |
→ |
m(ql(x0)) |
|
ql(0(x0)) |
→ |
0(ql(x0)) |
|
ql(1(x0)) |
→ |
1(ql(x0)) |
|
1(ql(b(x0))) |
→ |
r1(b(1(x0))) |
1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [r1(x1)] |
= |
1 ·
x1 + 0 |
| [r0(x1)] |
= |
2 ·
x1 + 3 |
| [m(x1)] |
= |
2 ·
x1 + 0 |
| [b(x1)] |
= |
2 ·
x1 + 0 |
| [0(x1)] |
= |
8 ·
x1 + 0 |
| [ql(x1)] |
= |
1 ·
x1 + 0 |
| [qr(x1)] |
= |
1 ·
x1 + 0 |
| [1(x1)] |
= |
1 ·
x1 + 0 |
the
rules
|
1(r0(x0)) |
→ |
r0(1(x0)) |
|
0(r1(x0)) |
→ |
r1(0(x0)) |
|
1(r1(x0)) |
→ |
r1(1(x0)) |
|
m(r1(x0)) |
→ |
r1(m(x0)) |
|
b(r1(x0)) |
→ |
b(1(qr(x0))) |
|
qr(0(x0)) |
→ |
0(qr(x0)) |
|
qr(1(x0)) |
→ |
1(qr(x0)) |
|
qr(m(x0)) |
→ |
m(ql(x0)) |
|
ql(0(x0)) |
→ |
0(ql(x0)) |
|
ql(1(x0)) |
→ |
1(ql(x0)) |
|
1(ql(b(x0))) |
→ |
r1(b(1(x0))) |
remain.
1.1.1.1.1.1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(0(x0)) |
→ |
0(r1(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
m(qr(x0)) |
→ |
ql(m(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
|
b(ql(1(x0))) |
→ |
1(b(r1(x0))) |
1.1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [r1(x1)] |
= |
·
x1 +
|
| [r0(x1)] |
= |
·
x1 +
|
| [m(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [0(x1)] |
= |
·
x1 +
|
| [ql(x1)] |
= |
·
x1 +
|
| [qr(x1)] |
= |
·
x1 +
|
| [1(x1)] |
= |
·
x1 +
|
the
rules
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
|
b(ql(1(x0))) |
→ |
1(b(r1(x0))) |
remain.
1.1.1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [r1(x1)] |
= |
0 ·
x1 +
-∞
|
| [r0(x1)] |
= |
4 ·
x1 +
-∞
|
| [m(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
4 ·
x1 +
-∞
|
| [0(x1)] |
= |
11 ·
x1 +
-∞
|
| [ql(x1)] |
= |
9 ·
x1 +
-∞
|
| [qr(x1)] |
= |
0 ·
x1 +
-∞
|
| [1(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
r1(b(x0)) |
→ |
qr(1(b(x0))) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
remain.
1.1.1.1.1.1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [r1(x1)] |
= |
8 ·
x1 +
-∞
|
| [r0(x1)] |
= |
8 ·
x1 +
-∞
|
| [m(x1)] |
= |
8 ·
x1 +
-∞
|
| [b(x1)] |
= |
3 ·
x1 +
-∞
|
| [0(x1)] |
= |
0 ·
x1 +
-∞
|
| [ql(x1)] |
= |
8 ·
x1 +
-∞
|
| [qr(x1)] |
= |
3 ·
x1 +
-∞
|
| [1(x1)] |
= |
1 ·
x1 +
-∞
|
the
rules
|
r0(1(x0)) |
→ |
1(r0(x0)) |
|
r1(1(x0)) |
→ |
1(r1(x0)) |
|
r1(m(x0)) |
→ |
m(r1(x0)) |
|
0(qr(x0)) |
→ |
qr(0(x0)) |
|
1(qr(x0)) |
→ |
qr(1(x0)) |
|
0(ql(x0)) |
→ |
ql(0(x0)) |
|
1(ql(x0)) |
→ |
ql(1(x0)) |
remain.
1.1.1.1.1.1.1.1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(ql) |
= |
0 |
|
weight(ql) |
= |
1 |
|
|
|
| prec(qr) |
= |
0 |
|
weight(qr) |
= |
1 |
|
|
|
| prec(r1) |
= |
3 |
|
weight(r1) |
= |
1 |
|
|
|
| prec(m) |
= |
0 |
|
weight(m) |
= |
1 |
|
|
|
| prec(1) |
= |
2 |
|
weight(1) |
= |
1 |
|
|
|
| prec(r0) |
= |
7 |
|
weight(r0) |
= |
0 |
|
|
|
| prec(0) |
= |
1 |
|
weight(0) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1.1.1.1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.