Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
xor(x,F) |
→ x |
| 2: |
|
xor(x,neg(x)) |
→ F |
| 3: |
|
and(x,T) |
→ x |
| 4: |
|
and(x,F) |
→ F |
| 5: |
|
and(x,x) |
→ x |
| 6: |
|
and(xor(x,y),z) |
→ xor(and(x,z),and(y,z)) |
| 7: |
|
xor(x,x) |
→ F |
| 8: |
|
impl(x,y) |
→ xor(and(x,y),xor(x,T)) |
| 9: |
|
or(x,y) |
→ xor(and(x,y),xor(x,y)) |
| 10: |
|
equiv(x,y) |
→ xor(x,xor(y,T)) |
| 11: |
|
neg(x) |
→ xor(x,T) |
|
There are 12 dependency pairs:
|
| 12: |
|
AND(xor(x,y),z) |
→ XOR(and(x,z),and(y,z)) |
| 13: |
|
AND(xor(x,y),z) |
→ AND(x,z) |
| 14: |
|
AND(xor(x,y),z) |
→ AND(y,z) |
| 15: |
|
IMPL(x,y) |
→ XOR(and(x,y),xor(x,T)) |
| 16: |
|
IMPL(x,y) |
→ AND(x,y) |
| 17: |
|
IMPL(x,y) |
→ XOR(x,T) |
| 18: |
|
OR(x,y) |
→ XOR(and(x,y),xor(x,y)) |
| 19: |
|
OR(x,y) |
→ AND(x,y) |
| 20: |
|
OR(x,y) |
→ XOR(x,y) |
| 21: |
|
EQUIV(x,y) |
→ XOR(x,xor(y,T)) |
| 22: |
|
EQUIV(x,y) |
→ XOR(y,T) |
| 23: |
|
NEG(x) |
→ XOR(x,T) |
|
The approximated dependency graph contains one SCC:
{13,14}.
-
Consider the SCC {13,14}.
There are no usable rules.
By taking the AF π with
π(AND) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {13,14}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.00 seconds)
--- May 3, 2006