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