Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
rev(nil) |
→ nil |
| 2: |
|
rev(rev(x)) |
→ x |
| 3: |
|
rev(x ++ y) |
→ rev(y) ++ rev(x) |
| 4: |
|
nil ++ y |
→ y |
| 5: |
|
x ++ nil |
→ x |
| 6: |
|
(x . y) ++ z |
→ x . (y ++ z) |
| 7: |
|
x ++ (y ++ z) |
→ (x ++ y) ++ z |
| 8: |
|
make(x) |
→ x . nil |
|
There are 6 dependency pairs:
|
| 9: |
|
REV(x ++ y) |
→ rev(y) ++# rev(x) |
| 10: |
|
REV(x ++ y) |
→ REV(y) |
| 11: |
|
REV(x ++ y) |
→ REV(x) |
| 12: |
|
(x . y) ++# z |
→ y ++# z |
| 13: |
|
x ++# (y ++ z) |
→ (x ++ y) ++# z |
| 14: |
|
x ++# (y ++ z) |
→ x ++# y |
|
The approximated dependency graph contains 2 SCCs:
{12-14}
and {10,11}.
-
Consider the SCC {12-14}.
The usable rules are {4-7}.
The constraints could not be solved.
-
Consider the SCC {10,11}.
There are no usable rules.
By taking the AF π with
π(REV) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {10,11}
are strictly decreasing.
Tyrolean Termination Tool (0.01 seconds)
--- May 4, 2006