Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
x + 0 |
→ x |
| 2: |
|
x + i(x) |
→ 0 |
| 3: |
|
(x + y) + z |
→ x + (y + z) |
| 4: |
|
x * (y + z) |
→ (x * y) + (x * z) |
| 5: |
|
(x + y) * z |
→ (x * z) + (y * z) |
|
There are 8 dependency pairs:
|
| 6: |
|
(x + y) +# z |
→ x +# (y + z) |
| 7: |
|
(x + y) +# z |
→ y +# z |
| 8: |
|
x *# (y + z) |
→ (x * y) +# (x * z) |
| 9: |
|
x *# (y + z) |
→ x *# y |
| 10: |
|
x *# (y + z) |
→ x *# z |
| 11: |
|
(x + y) *# z |
→ (x * z) +# (y * z) |
| 12: |
|
(x + y) *# z |
→ x *# z |
| 13: |
|
(x + y) *# z |
→ y *# z |
|
The approximated dependency graph contains 2 SCCs:
{6,7}
and {9,10,12,13}.
-
Consider the SCC {6,7}.
The usable rules are {1-3}.
By taking the AF π with
π(+#) = π(i) = 1 together with
the lexicographic path order with
precedence + ≻ 0,
the rules in {1-3,6,7}
are strictly decreasing.
-
Consider the SCC {9,10,12,13}.
There are no usable rules.
By taking the AF π with
π(*#) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {9,10}
are weakly decreasing and
the rules in {12,13}
are strictly decreasing.
There is one new SCC.
-
Consider the SCC {9,10}.
By taking the AF π with
π(*#) = 2 together with
the lexicographic path order with
empty precedence,
the rules in {9,10}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.02 seconds)
--- May 4, 2006