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