YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
f(x0) |
→ |
n(c(c(x0))) |
|
c(f(x0)) |
→ |
f(c(c(x0))) |
|
c(c(x0)) |
→ |
c(x0) |
|
n(s(x0)) |
→ |
f(s(s(x0))) |
|
n(f(x0)) |
→ |
f(n(x0)) |
Proof
1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [c(x1)] |
= |
·
x1 +
|
| [s(x1)] |
= |
·
x1 +
|
| [f(x1)] |
= |
·
x1 +
|
| [n(x1)] |
= |
·
x1 +
|
the
rules
|
c(f(x0)) |
→ |
f(c(c(x0))) |
|
c(c(x0)) |
→ |
c(x0) |
|
n(s(x0)) |
→ |
f(s(s(x0))) |
|
n(f(x0)) |
→ |
f(n(x0)) |
remain.
1.1 Rule Removal
Using the
linear polynomial interpretation over the arctic semiring over the integers
| [c(x1)] |
= |
0 ·
x1 +
-∞
|
| [s(x1)] |
= |
0 ·
x1 +
-∞
|
| [f(x1)] |
= |
0 ·
x1 +
-∞
|
| [n(x1)] |
= |
2 ·
x1 +
-∞
|
the
rules
|
c(f(x0)) |
→ |
f(c(c(x0))) |
|
c(c(x0)) |
→ |
c(x0) |
|
n(f(x0)) |
→ |
f(n(x0)) |
remain.
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight function
| prec(n) |
= |
1 |
|
weight(n) |
= |
1 |
|
|
|
| prec(c) |
= |
3 |
|
weight(c) |
= |
0 |
|
|
|
| prec(f) |
= |
0 |
|
weight(f) |
= |
1 |
|
|
|
all rules could be removed.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.