Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
f(a,x) |
→ f(b,f(c,x)) |
| 2: |
|
f(a,f(b,x)) |
→ f(b,f(a,x)) |
| 3: |
|
f(d,f(c,x)) |
→ f(d,f(a,x)) |
| 4: |
|
f(a,f(c,x)) |
→ f(c,f(a,x)) |
|
There are 8 dependency pairs:
|
| 5: |
|
F(a,x) |
→ F(b,f(c,x)) |
| 6: |
|
F(a,x) |
→ F(c,x) |
| 7: |
|
F(a,f(b,x)) |
→ F(b,f(a,x)) |
| 8: |
|
F(a,f(b,x)) |
→ F(a,x) |
| 9: |
|
F(d,f(c,x)) |
→ F(d,f(a,x)) |
| 10: |
|
F(d,f(c,x)) |
→ F(a,x) |
| 11: |
|
F(a,f(c,x)) |
→ F(c,f(a,x)) |
| 12: |
|
F(a,f(c,x)) |
→ F(a,x) |
|
The approximated dependency graph contains 2 SCCs:
{8,12}
and {9}.
-
Consider the SCC {8,12}.
There are no usable rules.
By taking the AF π with
π(F) = 2
and π(f) = [2] together with
the lexicographic path order with
empty precedence,
the rules in {8,12}
are strictly decreasing.
-
Consider the SCC {9}.
The constraints could not be solved.
Tyrolean Termination Tool (0.02 seconds)
--- May 3, 2006