Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
f(x,y,w,w,a) |
→ g1(x,x,y,w) |
| 2: |
|
f(x,y,w,a,a) |
→ g1(y,x,x,w) |
| 3: |
|
f(x,y,a,a,w) |
→ g2(x,y,y,w) |
| 4: |
|
f(x,y,a,w,w) |
→ g2(y,y,x,w) |
| 5: |
|
g1(x,x,y,a) |
→ h(x,y) |
| 6: |
|
g1(y,x,x,a) |
→ h(x,y) |
| 7: |
|
g2(x,y,y,a) |
→ h(x,y) |
| 8: |
|
g2(y,y,x,a) |
→ h(x,y) |
| 9: |
|
h(x,x) |
→ x |
|
There are 8 dependency pairs:
|
| 10: |
|
F(x,y,w,w,a) |
→ G1(x,x,y,w) |
| 11: |
|
F(x,y,w,a,a) |
→ G1(y,x,x,w) |
| 12: |
|
F(x,y,a,a,w) |
→ G2(x,y,y,w) |
| 13: |
|
F(x,y,a,w,w) |
→ G2(y,y,x,w) |
| 14: |
|
G1(x,x,y,a) |
→ H(x,y) |
| 15: |
|
G1(y,x,x,a) |
→ H(x,y) |
| 16: |
|
G2(x,y,y,a) |
→ H(x,y) |
| 17: |
|
G2(y,y,x,a) |
→ H(x,y) |
|
The approximated dependency graph contains no SCCs
and hence the TRS is trivially terminating.
Tyrolean Termination Tool (0.00 seconds)
--- May 4, 2006