Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
x + 0 |
→ x |
| 2: |
|
0 + x |
→ x |
| 3: |
|
s(x) + s(y) |
→ s(s(x + y)) |
| 4: |
|
x * 0 |
→ 0 |
| 5: |
|
0 * x |
→ 0 |
| 6: |
|
s(x) * s(y) |
→ s((x * y) + (x + y)) |
| 7: |
|
sum(nil) |
→ 0 |
| 8: |
|
sum(cons(x,l)) |
→ x + sum(l) |
| 9: |
|
prod(nil) |
→ s(0) |
| 10: |
|
prod(cons(x,l)) |
→ x * prod(l) |
|
There are 8 dependency pairs:
|
| 11: |
|
s(x) +# s(y) |
→ x +# y |
| 12: |
|
s(x) *# s(y) |
→ (x * y) +# (x + y) |
| 13: |
|
s(x) *# s(y) |
→ x *# y |
| 14: |
|
s(x) *# s(y) |
→ x +# y |
| 15: |
|
SUM(cons(x,l)) |
→ x +# sum(l) |
| 16: |
|
SUM(cons(x,l)) |
→ SUM(l) |
| 17: |
|
PROD(cons(x,l)) |
→ x *# prod(l) |
| 18: |
|
PROD(cons(x,l)) |
→ PROD(l) |
|
The approximated dependency graph contains 4 SCCs:
{11},
{13},
{18}
and {16}.
-
Consider the SCC {11}.
There are no usable rules.
By taking the AF π with
π(+#) = 1 together with
the lexicographic path order with
empty precedence,
rule 11
is strictly decreasing.
-
Consider the SCC {13}.
There are no usable rules.
By taking the AF π with
π(*#) = 1 together with
the lexicographic path order with
empty precedence,
rule 13
is strictly decreasing.
-
Consider the SCC {18}.
There are no usable rules.
By taking the AF π with
π(PROD) = 1
and π(cons) = [2] together with
the lexicographic path order with
empty precedence,
rule 18
is strictly decreasing.
-
Consider the SCC {16}.
There are no usable rules.
By taking the AF π with
π(SUM) = 1
and π(cons) = [2] together with
the lexicographic path order with
empty precedence,
rule 16
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.01 seconds)
--- May 4, 2006