Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
sqr(0) |
→ 0 |
| 2: |
|
sqr(s(x)) |
→ sqr(x) + s(double(x)) |
| 3: |
|
double(0) |
→ 0 |
| 4: |
|
double(s(x)) |
→ s(s(double(x))) |
| 5: |
|
x + 0 |
→ x |
| 6: |
|
x + s(y) |
→ s(x + y) |
| 7: |
|
sqr(s(x)) |
→ s(sqr(x) + double(x)) |
|
There are 6 dependency pairs:
|
| 8: |
|
SQR(s(x)) |
→ sqr(x) +# s(double(x)) |
| 9: |
|
DOUBLE(s(x)) |
→ DOUBLE(x) |
| 10: |
|
x +# s(y) |
→ x +# y |
| 11: |
|
SQR(s(x)) |
→ sqr(x) +# double(x) |
| 12: |
|
SQR(s(x)) |
→ SQR(x) |
| 13: |
|
SQR(s(x)) |
→ DOUBLE(x) |
|
The approximated dependency graph contains 3 SCCs:
{10},
{9}
and {12}.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(+#) = 2 together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
-
Consider the SCC {9}.
There are no usable rules.
By taking the AF π with
π(DOUBLE) = 1 together with
the lexicographic path order with
empty precedence,
rule 9
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 (0.01 seconds)
--- May 3, 2006