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