Conclusion
-1.
FORALL
A:finite
set[doubleton[T]],
S:set[doubleton[T]]
:
subset?(S,
A)
IMPLIES
is
finite(S)
1.
is
finite[doubleton[T]]({e:doubleton[T]
|
edges(G!1)(e)
AND
(FORALL
x:T
:
e(x)
IMPLIES
V!1(x))})
Tactic
INST?
Premise 1.   (has proof of 4 steps)
-1.
FORALL
A:finite
set[doubleton[T]]
:
subset?({e:doubleton[T]
|
edges(G!1)(e)
AND
(FORALL
x:T
:
e(x)
IMPLIES
V!1(x))},
A)
IMPLIES
is
finite({e:doubleton[T]
|
edges(G!1)(e)
AND
(FORALL
x:T
:
e(x)
IMPLIES
V!1(x))})
1.
is
finite[doubleton[T]]({e:doubleton[T]
|
edges(G!1)(e)
AND
(FORALL
x:T
:
e(x)
IMPLIES
V!1(x))})