Conclusion
1.
subset?({n:T
|
vert(G!1)(n)
AND
(n
/=
choose[T](vert(G!1)))
AND
(NOT
edge?[T](G!1)(choose[T](vert(G!1)),
n))},
vert(G!1))
Tactic
EXPAND
"subset?"
Premise 1.   (has proof of 3 steps)
1.
FORALL
x:T
:
member(x,
{n:T
|
vert(G!1)(n)
AND
(n
/=
choose[T](vert(G!1)))
AND
(NOT
edge?[T](G!1)(choose[T](vert(G!1)),
n))})
=
>
member(x,
vert(G!1))