Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
nats |
→ adx(zeros) |
| 2: |
|
zeros |
→ cons(0,zeros) |
| 3: |
|
incr(cons(X,Y)) |
→ cons(s(X),incr(Y)) |
| 4: |
|
adx(cons(X,Y)) |
→ incr(cons(X,adx(Y))) |
| 5: |
|
hd(cons(X,Y)) |
→ X |
| 6: |
|
tl(cons(X,Y)) |
→ Y |
|
There are 6 dependency pairs:
|
| 7: |
|
NATS |
→ ADX(zeros) |
| 8: |
|
NATS |
→ ZEROS |
| 9: |
|
ZEROS |
→ ZEROS |
| 10: |
|
INCR(cons(X,Y)) |
→ INCR(Y) |
| 11: |
|
ADX(cons(X,Y)) |
→ INCR(cons(X,adx(Y))) |
| 12: |
|
ADX(cons(X,Y)) |
→ ADX(Y) |
|
The approximated dependency graph contains 3 SCCs:
{10},
{12}
and {9}.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(INCR) = 1
and π(cons) = [2] together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
-
Consider the SCC {12}.
There are no usable rules.
By taking the AF π with
π(ADX) = 1
and π(cons) = [2] together with
the lexicographic path order with
empty precedence,
rule 12
is strictly decreasing.
-
Consider the SCC {9}.
There are no usable rules.
The constraints could not be solved.
Tyrolean Termination Tool (0.01 seconds)
--- May 4, 2006