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