Theorem sbfal

Description: Substitution does not change falsity. (Contributed by Giovanni Mascellani, 24-May-2019)

	Ref	Expression
Hypothesis	sbfal.1	|- A e. _V
Assertion	sbfal	|- ( [. A / x ]. F. <-> F. )

Proof

Step	Hyp	Ref	Expression
1		sbfal.1	|- A e. _V
2		sbcg	|- ( A e. _V -> ( [. A / x ]. F. <-> F. ) )
3	1 2	ax-mp	|- ( [. A / x ]. F. <-> F. )