Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
1: |
|
f(x,y) |
→ g1(x,x,y) |
2: |
|
f(x,y) |
→ g1(y,x,x) |
3: |
|
f(x,y) |
→ g2(x,y,y) |
4: |
|
f(x,y) |
→ g2(y,y,x) |
5: |
|
g1(x,x,y) |
→ h(x,y) |
6: |
|
g1(y,x,x) |
→ h(x,y) |
7: |
|
g2(x,y,y) |
→ h(x,y) |
8: |
|
g2(y,y,x) |
→ h(x,y) |
9: |
|
h(x,x) |
→ x |
|
There are 8 dependency pairs:
|
10: |
|
F(x,y) |
→ G1(x,x,y) |
11: |
|
F(x,y) |
→ G1(y,x,x) |
12: |
|
F(x,y) |
→ G2(x,y,y) |
13: |
|
F(x,y) |
→ G2(y,y,x) |
14: |
|
G1(x,x,y) |
→ H(x,y) |
15: |
|
G1(y,x,x) |
→ H(x,y) |
16: |
|
G2(x,y,y) |
→ H(x,y) |
17: |
|
G2(y,y,x) |
→ H(x,y) |
|
The approximated dependency graph contains no SCCs
and hence the TRS is trivially terminating.
Tyrolean Termination Tool (0.01 seconds)
--- May 4, 2006