Theorem bj-sbel1

Description: Version of sbcel1g when substituting a set. (Note: one could have a corresponding version of sbcel12 when substituting a set, but the point here is that the antecedent of sbcel1g is not needed when substituting a set.) (Contributed by BJ, 6-Oct-2018)

	Ref	Expression
Assertion	bj-sbel1	⊢ ( [ 𝑦 / 𝑥 ] 𝐴 ∈ 𝐵 ↔ ⦋ 𝑦 / 𝑥 ⦌ 𝐴 ∈ 𝐵 )

Proof

Step	Hyp	Ref	Expression
1		sbsbc	⊢ ( [ 𝑦 / 𝑥 ] 𝐴 ∈ 𝐵 ↔ [ 𝑦 / 𝑥 ] 𝐴 ∈ 𝐵 )
2		sbcel1g	⊢ ( 𝑦 ∈ V → ( [ 𝑦 / 𝑥 ] 𝐴 ∈ 𝐵 ↔ ⦋ 𝑦 / 𝑥 ⦌ 𝐴 ∈ 𝐵 ) )
3	2	elv	⊢ ( [ 𝑦 / 𝑥 ] 𝐴 ∈ 𝐵 ↔ ⦋ 𝑦 / 𝑥 ⦌ 𝐴 ∈ 𝐵 )
4	1 3	bitri	⊢ ( [ 𝑦 / 𝑥 ] 𝐴 ∈ 𝐵 ↔ ⦋ 𝑦 / 𝑥 ⦌ 𝐴 ∈ 𝐵 )