YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
r(r(x0)) |
→ |
s(r(x0)) |
|
r(s(x0)) |
→ |
s(r(x0)) |
|
r(n(x0)) |
→ |
s(r(x0)) |
|
r(b(x0)) |
→ |
u(s(b(x0))) |
|
r(u(x0)) |
→ |
u(r(x0)) |
|
s(u(x0)) |
→ |
u(s(x0)) |
|
n(u(x0)) |
→ |
u(n(x0)) |
|
t(r(u(x0))) |
→ |
t(c(r(x0))) |
|
t(s(u(x0))) |
→ |
t(c(r(x0))) |
|
t(n(u(x0))) |
→ |
t(c(r(x0))) |
|
c(u(x0)) |
→ |
u(c(x0)) |
|
c(s(x0)) |
→ |
s(c(x0)) |
|
c(r(x0)) |
→ |
r(c(x0)) |
|
c(n(x0)) |
→ |
n(c(x0)) |
|
c(n(x0)) |
→ |
n(x0) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [u(x1)] |
= |
0 ·
x1 +
-∞
|
| [s(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
0 ·
x1 +
-∞
|
| [t(x1)] |
= |
1 ·
x1 +
-∞
|
| [r(x1)] |
= |
0 ·
x1 +
-∞
|
| [c(x1)] |
= |
0 ·
x1 +
-∞
|
| [n(x1)] |
= |
9 ·
x1 +
-∞
|
the
rules
|
r(r(x0)) |
→ |
s(r(x0)) |
|
r(s(x0)) |
→ |
s(r(x0)) |
|
r(b(x0)) |
→ |
u(s(b(x0))) |
|
r(u(x0)) |
→ |
u(r(x0)) |
|
s(u(x0)) |
→ |
u(s(x0)) |
|
n(u(x0)) |
→ |
u(n(x0)) |
|
t(r(u(x0))) |
→ |
t(c(r(x0))) |
|
t(s(u(x0))) |
→ |
t(c(r(x0))) |
|
c(u(x0)) |
→ |
u(c(x0)) |
|
c(s(x0)) |
→ |
s(c(x0)) |
|
c(r(x0)) |
→ |
r(c(x0)) |
|
c(n(x0)) |
→ |
n(c(x0)) |
|
c(n(x0)) |
→ |
n(x0) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [u(x1)] |
= |
2 ·
x1 +
-∞
|
| [s(x1)] |
= |
0 ·
x1 +
-∞
|
| [b(x1)] |
= |
4 ·
x1 +
-∞
|
| [t(x1)] |
= |
0 ·
x1 +
-∞
|
| [r(x1)] |
= |
2 ·
x1 +
-∞
|
| [c(x1)] |
= |
0 ·
x1 +
-∞
|
| [n(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
r(s(x0)) |
→ |
s(r(x0)) |
|
r(b(x0)) |
→ |
u(s(b(x0))) |
|
r(u(x0)) |
→ |
u(r(x0)) |
|
s(u(x0)) |
→ |
u(s(x0)) |
|
n(u(x0)) |
→ |
u(n(x0)) |
|
t(s(u(x0))) |
→ |
t(c(r(x0))) |
|
c(u(x0)) |
→ |
u(c(x0)) |
|
c(s(x0)) |
→ |
s(c(x0)) |
|
c(r(x0)) |
→ |
r(c(x0)) |
|
c(n(x0)) |
→ |
n(c(x0)) |
|
c(n(x0)) |
→ |
n(x0) |
remain.
1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
s(r(x0)) |
→ |
r(s(x0)) |
|
b(r(x0)) |
→ |
b(s(u(x0))) |
|
u(r(x0)) |
→ |
r(u(x0)) |
|
u(s(x0)) |
→ |
s(u(x0)) |
|
u(n(x0)) |
→ |
n(u(x0)) |
|
u(s(t(x0))) |
→ |
r(c(t(x0))) |
|
u(c(x0)) |
→ |
c(u(x0)) |
|
s(c(x0)) |
→ |
c(s(x0)) |
|
r(c(x0)) |
→ |
c(r(x0)) |
|
n(c(x0)) |
→ |
c(n(x0)) |
|
n(c(x0)) |
→ |
n(x0) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the naturals
| [u(x1)] |
= |
·
x1 +
|
| [s(x1)] |
= |
·
x1 +
|
| [b(x1)] |
= |
·
x1 +
|
| [t(x1)] |
= |
·
x1 +
|
| [r(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
| [n(x1)] |
= |
·
x1 +
|
the
rules
|
s(r(x0)) |
→ |
r(s(x0)) |
|
u(r(x0)) |
→ |
r(u(x0)) |
|
u(s(x0)) |
→ |
s(u(x0)) |
|
u(n(x0)) |
→ |
n(u(x0)) |
|
u(s(t(x0))) |
→ |
r(c(t(x0))) |
|
u(c(x0)) |
→ |
c(u(x0)) |
|
s(c(x0)) |
→ |
c(s(x0)) |
|
r(c(x0)) |
→ |
c(r(x0)) |
|
n(c(x0)) |
→ |
c(n(x0)) |
|
n(c(x0)) |
→ |
n(x0) |
remain.
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [u(x1)] |
= |
14 ·
x1 +
-∞
|
| [s(x1)] |
= |
0 ·
x1 +
-∞
|
| [t(x1)] |
= |
4 ·
x1 +
-∞
|
| [r(x1)] |
= |
0 ·
x1 +
-∞
|
| [c(x1)] |
= |
14 ·
x1 +
-∞
|
| [n(x1)] |
= |
0 ·
x1 +
-∞
|
the
rules
|
s(r(x0)) |
→ |
r(s(x0)) |
|
u(r(x0)) |
→ |
r(u(x0)) |
|
u(s(x0)) |
→ |
s(u(x0)) |
|
u(n(x0)) |
→ |
n(u(x0)) |
|
u(s(t(x0))) |
→ |
r(c(t(x0))) |
|
u(c(x0)) |
→ |
c(u(x0)) |
|
s(c(x0)) |
→ |
c(s(x0)) |
|
r(c(x0)) |
→ |
c(r(x0)) |
|
n(c(x0)) |
→ |
c(n(x0)) |
remain.
1.1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(c) |
= |
0 |
|
weight(c) |
= |
1 |
|
|
|
| prec(t) |
= |
0 |
|
weight(t) |
= |
1 |
|
|
|
| prec(u) |
= |
3 |
|
weight(u) |
= |
1 |
|
|
|
| prec(n) |
= |
2 |
|
weight(n) |
= |
1 |
|
|
|
| prec(s) |
= |
2 |
|
weight(s) |
= |
1 |
|
|
|
| prec(r) |
= |
1 |
|
weight(r) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.