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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