Theorem sbcel2gv 3115

Description: Class substitution into a membership relation. (Contributed by NM, 17-Nov-2006.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)

Assertion

Ref	Expression
sbcel2gv	⊢ (𝐵 ∈ 𝑉 → ([𝐵 / 𝑥]𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵))

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝐵(𝑥)   𝑉(𝑥)

Proof of Theorem sbcel2gv

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq2 2302	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝑦))
2		eleq2 2302	. 2 ⊢ (𝑦 = 𝐵 → (𝐴 ∈ 𝑦 ↔ 𝐴 ∈ 𝐵))
3	1, 2	sbcie2g 3085	1 ⊢ (𝐵 ∈ 𝑉 → ([𝐵 / 𝑥]𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   ∈ wcel 2209  [wsbc 3051

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-sbc 3052

This theorem is referenced by:  sbcel21v  3116  csbvarg  3175