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 + y) + z | → x + (y + z) | 
| 5: |  | x * 0 | → 0 | 
| 6: |  | 0 * x | → 0 | 
| 7: |  | s(x) * s(y) | → s((x * y) + (x + y)) | 
| 8: |  | (x * y) * z | → x * (y * z) | 
| 9: |  | sum(nil) | → 0 | 
| 10: |  | sum(cons(x,l)) | → x + sum(l) | 
| 11: |  | prod(nil) | → s(0) | 
| 12: |  | prod(cons(x,l)) | → x * prod(l) | 
|  | 
There are 12 dependency pairs:
|  | 
| 13: |  | s(x) +# s(y) | → x +# y | 
| 14: |  | (x + y) +# z | → x +# (y + z) | 
| 15: |  | (x + y) +# z | → y +# z | 
| 16: |  | s(x) *# s(y) | → (x * y) +# (x + y) | 
| 17: |  | s(x) *# s(y) | → x *# y | 
| 18: |  | s(x) *# s(y) | → x +# y | 
| 19: |  | (x * y) *# z | → x *# (y * z) | 
| 20: |  | (x * y) *# z | → y *# z | 
| 21: |  | SUM(cons(x,l)) | → x +# sum(l) | 
| 22: |  | SUM(cons(x,l)) | → SUM(l) | 
| 23: |  | PROD(cons(x,l)) | → x *# prod(l) | 
| 24: |  | PROD(cons(x,l)) | → PROD(l) | 
|  | 
The approximated dependency graph contains 4 SCCs:
{13-15},
{17,19,20},
{24}
and {22}.
- 
Consider the SCC {13-15}.
The usable rules are {1-4}.
By taking the AF π with
π(+#) = π(s) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {3,13}
are weakly decreasing and
the rules in {1,2,4,14,15}
are strictly decreasing.
There is one new SCC.
- 
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 {17,19,20}.
The usable rules are {1-8}.
By taking the AF π with
π(*#) = 1 together with
the lexicographic path order with
precedence * ≻ + ≻ s,
the rules in {1-8,17,19,20}
are strictly decreasing.
- 
Consider the SCC {24}.
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 24
is strictly decreasing.
- 
Consider the SCC {22}.
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 22
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool  (0.03 seconds)
  ---  May 4, 2006