YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
0(P(x0)) |
→ |
1(P(x0)) |
|
1(P(x0)) |
→ |
c(P(x0)) |
|
0(c(x0)) |
→ |
1(0(x0)) |
|
1(c(x0)) |
→ |
c(0(x0)) |
|
P(0(x0)) |
→ |
P(1(0(0(x0)))) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1
over the arctic semiring over the integers
| [0(x1)] |
= |
·
x1 +
|
| [c(x1)] |
= |
·
x1 +
|
| [P(x1)] |
= |
·
x1 +
|
| [1(x1)] |
= |
·
x1 +
|
the
rules
|
0(P(x0)) |
→ |
1(P(x0)) |
|
1(P(x0)) |
→ |
c(P(x0)) |
|
0(c(x0)) |
→ |
1(0(x0)) |
|
1(c(x0)) |
→ |
c(0(x0)) |
remain.
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(1) |
= |
2 |
|
weight(1) |
= |
1 |
|
|
|
| prec(0) |
= |
3 |
|
weight(0) |
= |
1 |
|
|
|
| prec(P) |
= |
0 |
|
weight(P) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.