Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
f(X) |
→ cons(X,n__f(g(X))) |
| 2: |
|
g(0) |
→ s(0) |
| 3: |
|
g(s(X)) |
→ s(s(g(X))) |
| 4: |
|
sel(0,cons(X,Y)) |
→ X |
| 5: |
|
sel(s(X),cons(Y,Z)) |
→ sel(X,activate(Z)) |
| 6: |
|
f(X) |
→ n__f(X) |
| 7: |
|
activate(n__f(X)) |
→ f(X) |
| 8: |
|
activate(X) |
→ X |
|
There are 5 dependency pairs:
|
| 9: |
|
F(X) |
→ G(X) |
| 10: |
|
G(s(X)) |
→ G(X) |
| 11: |
|
SEL(s(X),cons(Y,Z)) |
→ SEL(X,activate(Z)) |
| 12: |
|
SEL(s(X),cons(Y,Z)) |
→ ACTIVATE(Z) |
| 13: |
|
ACTIVATE(n__f(X)) |
→ F(X) |
|
The approximated dependency graph contains 2 SCCs:
{10}
and {11}.
-
Consider the SCC {10}.
There are no usable rules.
By taking the AF π with
π(G) = 1 together with
the lexicographic path order with
empty precedence,
rule 10
is strictly decreasing.
-
Consider the SCC {11}.
The usable rules are {1-3,6-8}.
By taking the AF π with
π(activate) = π(cons) = π(f) = π(n__f) = π(SEL) = 1 together with
the lexicographic path order with
precedence g ≻ s,
the rules in {1,6-8}
are weakly decreasing and
the rules in {2,3,11}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.05 seconds)
--- May 4, 2006