Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
i(0) |
→ 0 |
| 2: |
|
0 + y |
→ y |
| 3: |
|
x + 0 |
→ x |
| 4: |
|
i(i(x)) |
→ x |
| 5: |
|
i(x) + x |
→ 0 |
| 6: |
|
x + i(x) |
→ 0 |
| 7: |
|
i(x + y) |
→ i(x) + i(y) |
| 8: |
|
x + (y + z) |
→ (x + y) + z |
| 9: |
|
(x + i(y)) + y |
→ x |
| 10: |
|
(x + y) + i(y) |
→ x |
|
There are 5 dependency pairs:
|
| 11: |
|
I(x + y) |
→ i(x) +# i(y) |
| 12: |
|
I(x + y) |
→ I(x) |
| 13: |
|
I(x + y) |
→ I(y) |
| 14: |
|
x +# (y + z) |
→ (x + y) +# z |
| 15: |
|
x +# (y + z) |
→ x +# y |
|
The approximated dependency graph contains 2 SCCs:
{14,15}
and {12,13}.
-
Consider the SCC {14,15}.
The usable rules are {2,3,5,6,8-10}.
The constraints could not be solved.
-
Consider the SCC {12,13}.
There are no usable rules.
By taking the AF π with
π(I) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {12,13}
are strictly decreasing.
Tyrolean Termination Tool (0.02 seconds)
--- May 4, 2006