Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
x + 0 |
→ x |
| 2: |
|
minus(x) + x |
→ 0 |
| 3: |
|
minus(0) |
→ 0 |
| 4: |
|
minus(minus(x)) |
→ x |
| 5: |
|
minus(x + y) |
→ minus(y) + minus(x) |
| 6: |
|
x * 1 |
→ x |
| 7: |
|
x * 0 |
→ 0 |
| 8: |
|
x * (y + z) |
→ (x * y) + (x * z) |
| 9: |
|
x * minus(y) |
→ minus(x * y) |
|
There are 8 dependency pairs:
|
| 10: |
|
MINUS(x + y) |
→ minus(y) +# minus(x) |
| 11: |
|
MINUS(x + y) |
→ MINUS(y) |
| 12: |
|
MINUS(x + y) |
→ MINUS(x) |
| 13: |
|
x *# (y + z) |
→ (x * y) +# (x * z) |
| 14: |
|
x *# (y + z) |
→ x *# y |
| 15: |
|
x *# (y + z) |
→ x *# z |
| 16: |
|
x *# minus(y) |
→ MINUS(x * y) |
| 17: |
|
x *# minus(y) |
→ x *# y |
|
The approximated dependency graph contains 2 SCCs:
{11,12}
and {14,15,17}.
-
Consider the SCC {11,12}.
There are no usable rules.
By taking the AF π with
π(MINUS) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {11,12}
are strictly decreasing.
-
Consider the SCC {14,15,17}.
There are no usable rules.
By taking the AF π with
π(minus) = 1
and π(*#) = 2 together with
the lexicographic path order with
empty precedence,
rule 17
is weakly decreasing and
the rules in {14,15}
are strictly decreasing.
There is one new SCC.
-
Consider the SCC {17}.
By taking the AF π with
π(*#) = 2 together with
the lexicographic path order with
empty precedence,
rule 17
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.01 seconds)
--- May 3, 2006