Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
s(a) |
→ a |
| 2: |
|
s(s(x)) |
→ x |
| 3: |
|
s(f(x,y)) |
→ f(s(y),s(x)) |
| 4: |
|
s(g(x,y)) |
→ g(s(x),s(y)) |
| 5: |
|
f(x,a) |
→ x |
| 6: |
|
f(a,y) |
→ y |
| 7: |
|
f(g(x,y),g(u,v)) |
→ g(f(x,u),f(y,v)) |
| 8: |
|
g(a,a) |
→ a |
|
There are 9 dependency pairs:
|
| 9: |
|
S(f(x,y)) |
→ F(s(y),s(x)) |
| 10: |
|
S(f(x,y)) |
→ S(y) |
| 11: |
|
S(f(x,y)) |
→ S(x) |
| 12: |
|
S(g(x,y)) |
→ G(s(x),s(y)) |
| 13: |
|
S(g(x,y)) |
→ S(x) |
| 14: |
|
S(g(x,y)) |
→ S(y) |
| 15: |
|
F(g(x,y),g(u,v)) |
→ G(f(x,u),f(y,v)) |
| 16: |
|
F(g(x,y),g(u,v)) |
→ F(x,u) |
| 17: |
|
F(g(x,y),g(u,v)) |
→ F(y,v) |
|
The approximated dependency graph contains 2 SCCs:
{16,17}
and {10,11,13,14}.
-
Consider the SCC {16,17}.
There are no usable rules.
By taking the AF π with
π(F) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {16,17}
are strictly decreasing.
-
Consider the SCC {10,11,13,14}.
There are no usable rules.
By taking the AF π with
π(S) = 1 together with
the lexicographic path order with
empty precedence,
the rules in {10,11,13,14}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.01 seconds)
--- May 4, 2006