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