Theorem sbfal

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

	Ref	Expression
Hypothesis	sbfal.1	⊢ 𝐴 ∈ V
Assertion	sbfal	⊢ ( [ 𝐴 / 𝑥 ] ⊥ ↔ ⊥ )

Proof

Step	Hyp	Ref	Expression
1		sbfal.1	⊢ 𝐴 ∈ V
2		sbcg	⊢ ( 𝐴 ∈ V → ( [ 𝐴 / 𝑥 ] ⊥ ↔ ⊥ ) )
3	1 2	ax-mp	⊢ ( [ 𝐴 / 𝑥 ] ⊥ ↔ ⊥ )