Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
eq(0,0) |
→ true |
| 2: |
|
eq(s(X),s(Y)) |
→ eq(X,Y) |
| 3: |
|
eq(X,Y) |
→ false |
| 4: |
|
inf(X) |
→ cons(X,inf(s(X))) |
| 5: |
|
take(0,X) |
→ nil |
| 6: |
|
take(s(X),cons(Y,L)) |
→ cons(Y,take(X,L)) |
| 7: |
|
length(nil) |
→ 0 |
| 8: |
|
length(cons(X,L)) |
→ s(length(L)) |
|
There are 4 dependency pairs:
|
| 9: |
|
EQ(s(X),s(Y)) |
→ EQ(X,Y) |
| 10: |
|
INF(X) |
→ INF(s(X)) |
| 11: |
|
TAKE(s(X),cons(Y,L)) |
→ TAKE(X,L) |
| 12: |
|
LENGTH(cons(X,L)) |
→ LENGTH(L) |
|
The approximated dependency graph contains 4 SCCs:
{9},
{10},
{12}
and {11}.
-
Consider the SCC {9}.
There are no usable rules.
By taking the AF π with
π(EQ) = 1 together with
the lexicographic path order with
empty precedence,
rule 9
is strictly decreasing.
-
Consider the SCC {10}.
There are no usable rules.
The constraints could not be solved.
-
Consider the SCC {12}.
There are no usable rules.
By taking the AF π with
π(LENGTH) = 1
and π(cons) = [2] together with
the lexicographic path order with
empty precedence,
rule 12
is strictly decreasing.
-
Consider the SCC {11}.
There are no usable rules.
By taking the AF π with
π(TAKE) = 1 together with
the lexicographic path order with
empty precedence,
rule 11
is strictly decreasing.
Tyrolean Termination Tool (0.01 seconds)
--- May 3, 2006