Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
terms(N) |
→ cons(recip(sqr(N))) |
| 2: |
|
sqr(0) |
→ 0 |
| 3: |
|
sqr(s(X)) |
→ s(add(sqr(X),dbl(X))) |
| 4: |
|
dbl(0) |
→ 0 |
| 5: |
|
dbl(s(X)) |
→ s(s(dbl(X))) |
| 6: |
|
add(0,X) |
→ X |
| 7: |
|
add(s(X),Y) |
→ s(add(X,Y)) |
| 8: |
|
first(0,X) |
→ nil |
| 9: |
|
first(s(X),cons(Y)) |
→ cons(Y) |
|
There are 6 dependency pairs:
|
| 10: |
|
TERMS(N) |
→ SQR(N) |
| 11: |
|
SQR(s(X)) |
→ ADD(sqr(X),dbl(X)) |
| 12: |
|
SQR(s(X)) |
→ SQR(X) |
| 13: |
|
SQR(s(X)) |
→ DBL(X) |
| 14: |
|
DBL(s(X)) |
→ DBL(X) |
| 15: |
|
ADD(s(X),Y) |
→ ADD(X,Y) |
|
The approximated dependency graph contains 3 SCCs:
{15},
{14}
and {12}.
-
Consider the SCC {15}.
There are no usable rules.
By taking the AF π with
π(ADD) = 1 together with
the lexicographic path order with
empty precedence,
rule 15
is strictly decreasing.
-
Consider the SCC {14}.
There are no usable rules.
By taking the AF π with
π(DBL) = 1 together with
the lexicographic path order with
empty precedence,
rule 14
is strictly decreasing.
-
Consider the SCC {12}.
There are no usable rules.
By taking the AF π with
π(SQR) = 1 together with
the lexicographic path order with
empty precedence,
rule 12
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.02 seconds)
--- May 4, 2006