Theorem sbcel2gv 3805

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 2849	. 2 ⊢ (𝑥 = 𝑦 → (𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝑦))
2		eleq2 2849	. 2 ⊢ (𝑦 = 𝐵 → (𝐴 ∈ 𝑦 ↔ 𝐴 ∈ 𝐵))
3	1, 2	sbcie2g 3779	1 ⊢ (𝐵 ∈ 𝑉 → ([𝐵 / 𝑥]𝐴 ∈ 𝑥 ↔ 𝐴 ∈ 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∈ wcel 2145  [wsbc 3739

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3740

This theorem is used by:  sbcel21v  3806  csbvarg  4392  bnj92  35371  bnj539  35400  minregex  44374  frege77  44780  sbcoreleleq  45358  trsbc  45363  onfrALTlem5  45365  sbcoreleleqVD  45681  trsbcVD  45699  onfrALTlem5VD  45707  modelaxreplem3  45803