Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
and(x,or(y,z)) |
→ or(and(x,y),and(x,z)) |
| 2: |
|
and(x,and(y,y)) |
→ and(x,y) |
| 3: |
|
or(or(x,y),and(y,z)) |
→ or(x,y) |
| 4: |
|
or(x,and(x,y)) |
→ x |
| 5: |
|
or(true,y) |
→ true |
| 6: |
|
or(x,false) |
→ x |
| 7: |
|
or(x,x) |
→ x |
| 8: |
|
or(x,or(y,y)) |
→ or(x,y) |
| 9: |
|
and(x,true) |
→ x |
| 10: |
|
and(false,y) |
→ false |
| 11: |
|
and(x,x) |
→ x |
|
There are 5 dependency pairs:
|
| 12: |
|
AND(x,or(y,z)) |
→ OR(and(x,y),and(x,z)) |
| 13: |
|
AND(x,or(y,z)) |
→ AND(x,y) |
| 14: |
|
AND(x,or(y,z)) |
→ AND(x,z) |
| 15: |
|
AND(x,and(y,y)) |
→ AND(x,y) |
| 16: |
|
OR(x,or(y,y)) |
→ OR(x,y) |
|
The approximated dependency graph contains 2 SCCs:
{16}
and {13-15}.
-
Consider the SCC {16}.
There are no usable rules.
By taking the AF π with
π(OR) = 2
and π(or) = [1] together with
the lexicographic path order with
empty precedence,
rule 16
is strictly decreasing.
-
Consider the SCC {13-15}.
There are no usable rules.
By taking the AF π with
π(and) = 1
and π(AND) = 2 together with
the lexicographic path order with
empty precedence,
rule 15
is weakly decreasing and
the rules in {13,14}
are strictly decreasing.
There is one new SCC.
-
Consider the SCC {15}.
By taking the AF π with
π(AND) = 2
and π(and) = [1] together with
the lexicographic path order with
empty precedence,
rule 15
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.02 seconds)
--- May 3, 2006