Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
c(b(a(X))) |
→ a(a(b(b(c(c(X)))))) |
| 2: |
|
a(X) |
→ e |
| 3: |
|
b(X) |
→ e |
| 4: |
|
c(X) |
→ e |
|
There are 6 dependency pairs:
|
| 5: |
|
C(b(a(X))) |
→ A(a(b(b(c(c(X)))))) |
| 6: |
|
C(b(a(X))) |
→ A(b(b(c(c(X))))) |
| 7: |
|
C(b(a(X))) |
→ B(b(c(c(X)))) |
| 8: |
|
C(b(a(X))) |
→ B(c(c(X))) |
| 9: |
|
C(b(a(X))) |
→ C(c(X)) |
| 10: |
|
C(b(a(X))) |
→ C(X) |
|
The approximated dependency graph contains one SCC:
{10}.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(b) = π(C) = 1 together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.01 seconds)
--- May 3, 2006