Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
int(0,0) |
→ 0 . nil |
| 2: |
|
int(0,s(y)) |
→ 0 . int(s(0),s(y)) |
| 3: |
|
int(s(x),0) |
→ nil |
| 4: |
|
int(s(x),s(y)) |
→ int_list(int(x,y)) |
| 5: |
|
int_list(nil) |
→ nil |
| 6: |
|
int_list(x . y) |
→ s(x) . int_list(y) |
|
There are 4 dependency pairs:
|
| 7: |
|
INT(0,s(y)) |
→ INT(s(0),s(y)) |
| 8: |
|
INT(s(x),s(y)) |
→ INT_LIST(int(x,y)) |
| 9: |
|
INT(s(x),s(y)) |
→ INT(x,y) |
| 10: |
|
INT_LIST(x . y) |
→ INT_LIST(y) |
|
The approximated dependency graph contains 2 SCCs:
{10}
and {7,9}.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(INT_LIST) = 1
and π(.) = [2] together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
-
Consider the SCC {7,9}.
There are no usable rules.
By taking the AF π with
π(INT) = 2 together with
the lexicographic path order with
empty precedence,
rule 7
is weakly decreasing and
rule 9
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.02 seconds)
--- May 3, 2006