Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
fib(0) |
→ 0 |
| 2: |
|
fib(s(0)) |
→ s(0) |
| 3: |
|
fib(s(s(0))) |
→ s(0) |
| 4: |
|
fib(s(s(x))) |
→ sp(g(x)) |
| 5: |
|
g(0) |
→ pair(s(0),0) |
| 6: |
|
g(s(0)) |
→ pair(s(0),s(0)) |
| 7: |
|
g(s(x)) |
→ np(g(x)) |
| 8: |
|
sp(pair(x,y)) |
→ x + y |
| 9: |
|
np(pair(x,y)) |
→ pair(x + y,x) |
| 10: |
|
x + 0 |
→ x |
| 11: |
|
x + s(y) |
→ s(x + y) |
|
There are 7 dependency pairs:
|
| 12: |
|
FIB(s(s(x))) |
→ SP(g(x)) |
| 13: |
|
FIB(s(s(x))) |
→ G(x) |
| 14: |
|
G(s(x)) |
→ NP(g(x)) |
| 15: |
|
G(s(x)) |
→ G(x) |
| 16: |
|
SP(pair(x,y)) |
→ x +# y |
| 17: |
|
NP(pair(x,y)) |
→ x +# y |
| 18: |
|
x +# s(y) |
→ x +# y |
|
The approximated dependency graph contains 2 SCCs:
{18}
and {15}.
-
Consider the SCC {18}.
There are no usable rules.
By taking the AF π with
π(+#) = 2 together with
the lexicographic path order with
empty precedence,
rule 18
is strictly decreasing.
-
Consider the SCC {15}.
There are no usable rules.
By taking the AF π with
π(G) = 1 together with
the lexicographic path order with
empty precedence,
rule 15
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.00 seconds)
--- May 3, 2006