Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
f(f(x)) |
→ f(c(f(x))) |
| 2: |
|
f(f(x)) |
→ f(d(f(x))) |
| 3: |
|
g(c(x)) |
→ x |
| 4: |
|
g(d(x)) |
→ x |
| 5: |
|
g(c(h(0))) |
→ g(d(1)) |
| 6: |
|
g(c(1)) |
→ g(d(h(0))) |
| 7: |
|
g(h(x)) |
→ g(x) |
|
There are 5 dependency pairs:
|
| 8: |
|
F(f(x)) |
→ F(c(f(x))) |
| 9: |
|
F(f(x)) |
→ F(d(f(x))) |
| 10: |
|
G(c(h(0))) |
→ G(d(1)) |
| 11: |
|
G(c(1)) |
→ G(d(h(0))) |
| 12: |
|
G(h(x)) |
→ G(x) |
|
The approximated dependency graph contains one SCC:
{12}.
-
Consider the SCC {12}.
There are no usable rules.
By taking the AF π with
π(G) = 1 together with
the lexicographic path order with
empty precedence,
rule 12
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.00 seconds)
--- May 3, 2006