Conclusion
-1.
member(x!1,
{n:T
|
vert(G!1)(n)
AND
(n
/=
choose[T](vert(G!1)))
AND
edge?[T](G!1)(choose[T](vert(G!1)),
n)})
1.
member(x!1,
vert(G!1))
Tactic
EXPAND
"member"
Premise 1.   (has proof of 1 step)
-1.
vert(G!1)(x!1)
AND
(x!1
/=
choose[T](vert(G!1)))
AND
edge?[T](G!1)(choose[T](vert(G!1)),
x!1)
1.
vert(G!1)(x!1)