Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
p(0) |
→ s(s(0)) |
| 2: |
|
p(s(x)) |
→ x |
| 3: |
|
p(p(s(x))) |
→ p(x) |
| 4: |
|
le(p(s(x)),x) |
→ le(x,x) |
| 5: |
|
le(0,y) |
→ true |
| 6: |
|
le(s(x),0) |
→ false |
| 7: |
|
le(s(x),s(y)) |
→ le(x,y) |
| 8: |
|
minus(x,y) |
→ if(le(x,y),x,y) |
| 9: |
|
if(true,x,y) |
→ 0 |
| 10: |
|
if(false,x,y) |
→ s(minus(p(x),y)) |
|
There are 7 dependency pairs:
|
| 11: |
|
P(p(s(x))) |
→ P(x) |
| 12: |
|
LE(p(s(x)),x) |
→ LE(x,x) |
| 13: |
|
LE(s(x),s(y)) |
→ LE(x,y) |
| 14: |
|
MINUS(x,y) |
→ IF(le(x,y),x,y) |
| 15: |
|
MINUS(x,y) |
→ LE(x,y) |
| 16: |
|
IF(false,x,y) |
→ MINUS(p(x),y) |
| 17: |
|
IF(false,x,y) |
→ P(x) |
|
The approximated dependency graph contains 3 SCCs:
{12,13},
{11}
and {14,16}.
-
Consider the SCC {12,13}.
There are no usable rules.
By taking the AF π with
π(LE) = π(p) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {12,13}
are strictly decreasing.
-
Consider the SCC {11}.
There are no usable rules.
By taking the AF π with
π(p) = π(P) = 1 together with
the lexicographic path order with
empty precedence,
rule 11
is strictly decreasing.
-
Consider the SCC {14,16}.
The usable rules are {1-7}.
The constraints could not be solved.
Tyrolean Termination Tool (0.04 seconds)
--- May 3, 2006