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Metamath Proof Explorer

Theorem sbcel1g

Description: Move proper substitution in and out of a membership relation. Note that the scope of [. A / x ]. is the wff B e. C , whereas the scope of [_ A / x ]_ is the class B . (Contributed by NM, 10-Nov-2005)

		Ref	Expression
	Assertion	sbcel1g	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 ∈ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		sbcel12	⊢ ( [ 𝐴 / 𝑥 ] 𝐵 ∈ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 )
2		csbconstg	⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 𝐶 = 𝐶 )
3	2	eleq2d	⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ ⦋ 𝐴 / 𝑥 ⦌ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ 𝐶 ) )
4	1 3	bitrid	⊢ ( 𝐴 ∈ 𝑉 → ( [ 𝐴 / 𝑥 ] 𝐵 ∈ 𝐶 ↔ ⦋ 𝐴 / 𝑥 ⦌ 𝐵 ∈ 𝐶 ) )