Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
f(X) |
→ cons(X,n__f(n__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: |
|
g(X) |
→ n__g(X) |
| 8: |
|
activate(n__f(X)) |
→ f(activate(X)) |
| 9: |
|
activate(n__g(X)) |
→ g(activate(X)) |
| 10: |
|
activate(X) |
→ X |
|
There are 7 dependency pairs:
|
| 11: |
|
G(s(X)) |
→ G(X) |
| 12: |
|
SEL(s(X),cons(Y,Z)) |
→ SEL(X,activate(Z)) |
| 13: |
|
SEL(s(X),cons(Y,Z)) |
→ ACTIVATE(Z) |
| 14: |
|
ACTIVATE(n__f(X)) |
→ F(activate(X)) |
| 15: |
|
ACTIVATE(n__f(X)) |
→ ACTIVATE(X) |
| 16: |
|
ACTIVATE(n__g(X)) |
→ G(activate(X)) |
| 17: |
|
ACTIVATE(n__g(X)) |
→ ACTIVATE(X) |
|
The approximated dependency graph contains 3 SCCs:
{11},
{15,17}
and {12}.
-
Consider the SCC {11}.
There are no usable rules.
By taking the AF π with
π(G) = 1 together with
the lexicographic path order with
empty precedence,
rule 11
is strictly decreasing.
-
Consider the SCC {15,17}.
There are no usable rules.
By taking the AF π with
π(ACTIVATE) = π(n__f) = 1 together with
the lexicographic path order with
empty precedence,
rule 15
is weakly decreasing and
rule 17
is strictly decreasing.
There is one new SCC.
-
Consider the SCC {15}.
By taking the AF π with
π(ACTIVATE) = 1 together with
the lexicographic path order with
empty precedence,
rule 15
is strictly decreasing.
-
Consider the SCC {12}.
The usable rules are {1-3,6-10}.
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 n__g ≻ s
and n__g ≈ g,
the rules in {1,6-10}
are weakly decreasing and
the rules in {2,3,12}
are strictly decreasing.
Hence the TRS is terminating.
Tyrolean Termination Tool (0.09 seconds)
--- May 4, 2006