Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
minus(x,0) |
→ x |
| 2: |
|
minus(s(x),s(y)) |
→ minus(x,y) |
| 3: |
|
f(0) |
→ s(0) |
| 4: |
|
f(s(x)) |
→ minus(s(x),g(f(x))) |
| 5: |
|
g(0) |
→ 0 |
| 6: |
|
g(s(x)) |
→ minus(s(x),f(g(x))) |
|
There are 7 dependency pairs:
|
| 7: |
|
MINUS(s(x),s(y)) |
→ MINUS(x,y) |
| 8: |
|
F(s(x)) |
→ MINUS(s(x),g(f(x))) |
| 9: |
|
F(s(x)) |
→ G(f(x)) |
| 10: |
|
F(s(x)) |
→ F(x) |
| 11: |
|
G(s(x)) |
→ MINUS(s(x),f(g(x))) |
| 12: |
|
G(s(x)) |
→ F(g(x)) |
| 13: |
|
G(s(x)) |
→ G(x) |
|
The approximated dependency graph contains 2 SCCs:
{7}
and {9,10,12,13}.
-
Consider the SCC {7}.
There are no usable rules.
By taking the AF π with
π(MINUS) = 1 together with
the lexicographic path order with
empty precedence,
rule 7
is strictly decreasing.
-
Consider the SCC {9,10,12,13}.
By taking the AF π with
π(F) = π(g) = π(G) = π(minus) = 1 together with
the lexicographic path order with
precedence s ≈ f,
the rules in {1,3,5,6,9}
are weakly decreasing and
the rules in {2,4,10,12,13}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.02 seconds)
--- May 4, 2006