Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
x - 0 |
→ x |
| 2: |
|
0 - s(y) |
→ 0 |
| 3: |
|
s(x) - s(y) |
→ x - y |
| 4: |
|
f(0) |
→ 0 |
| 5: |
|
f(s(x)) |
→ s(x) - g(f(x)) |
| 6: |
|
g(0) |
→ s(0) |
| 7: |
|
g(s(x)) |
→ s(x) - f(g(x)) |
|
There are 7 dependency pairs:
|
| 8: |
|
s(x) -# s(y) |
→ x -# y |
| 9: |
|
F(s(x)) |
→ s(x) -# g(f(x)) |
| 10: |
|
F(s(x)) |
→ G(f(x)) |
| 11: |
|
F(s(x)) |
→ F(x) |
| 12: |
|
G(s(x)) |
→ s(x) -# f(g(x)) |
| 13: |
|
G(s(x)) |
→ F(g(x)) |
| 14: |
|
G(s(x)) |
→ G(x) |
|
The approximated dependency graph contains 2 SCCs:
{8}
and {10,11,13,14}.
-
Consider the SCC {8}.
There are no usable rules.
By taking the AF π with
π(-#) = 1 together with
the lexicographic path order with
empty precedence,
rule 8
is strictly decreasing.
-
Consider the SCC {10,11,13,14}.
By taking the AF π with
π(-) = π(f) = π(F) = π(G) = 1 together with
the lexicographic path order with
precedence s ≈ g,
the rules in {1,2,4-6,13}
are weakly decreasing and
the rules in {3,7,10,11,14}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.04 seconds)
--- May 4, 2006