Theorem sbcel1gv 2821

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

Assertion

Ref	Expression
sbcel1gv	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))

Distinct variable group:   𝑥,𝐵

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

Proof of Theorem sbcel1gv

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfsbcq2 2767	. 2 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝑥 ∈ 𝐵 ↔ [𝐴 / 𝑥]𝑥 ∈ 𝐵))
2		eleq1 2100	. 2 ⊢ (𝑦 = 𝐴 → (𝑦 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))
3		clelsb3 2142	. 2 ⊢ ([𝑦 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵)
4	1, 2, 3	vtoclbg 2614	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑥 ∈ 𝐵 ↔ 𝐴 ∈ 𝐵))

Syntax hints:   → wi 4   ↔ wb 98   ∈ wcel 1393  [wsb 1645  [wsbc 2764

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-sbc 2765

This theorem is referenced by: (None)