Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
fst(0,Z) |
→ nil |
| 2: |
|
fst(s(X),cons(Y,Z)) |
→ cons(Y,fst(X,Z)) |
| 3: |
|
from(X) |
→ cons(X,from(s(X))) |
| 4: |
|
add(0,X) |
→ X |
| 5: |
|
add(s(X),Y) |
→ s(add(X,Y)) |
| 6: |
|
len(nil) |
→ 0 |
| 7: |
|
len(cons(X,Z)) |
→ s(len(Z)) |
|
There are 4 dependency pairs:
|
| 8: |
|
FST(s(X),cons(Y,Z)) |
→ FST(X,Z) |
| 9: |
|
FROM(X) |
→ FROM(s(X)) |
| 10: |
|
ADD(s(X),Y) |
→ ADD(X,Y) |
| 11: |
|
LEN(cons(X,Z)) |
→ LEN(Z) |
|
The approximated dependency graph contains 4 SCCs:
{10},
{9},
{8}
and {11}.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(ADD) = 1 together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
-
Consider the SCC {9}.
There are no usable rules.
The constraints could not be solved.
-
Consider the SCC {8}.
There are no usable rules.
By taking the AF π with
π(FST) = 1 together with
the lexicographic path order with
empty precedence,
rule 8
is strictly decreasing.
-
Consider the SCC {11}.
There are no usable rules.
By taking the AF π with
π(LEN) = 1
and π(cons) = [2] together with
the lexicographic path order with
empty precedence,
rule 11
is strictly decreasing.
Tyrolean Termination Tool (0.01 seconds)
--- May 4, 2006