Theorem sbcel1g 3103

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

Assertion

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

Distinct variable group:   𝑥,𝐶

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

Proof of Theorem sbcel1g

Step	Hyp	Ref	Expression
1		sbcel12g 3099	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3098	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eleq2d 2266	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ 𝐶))
4	1, 3	bitrd 188	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ 𝐶))

Syntax hints:   → wi 4   ↔ wb 105   ∈ wcel 2167  [wsbc 2989  ⦋csb 3084

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 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-v 2765  df-sbc 2990  df-csb 3085

This theorem is referenced by:  rspcsbela  3144  fprodcllemf  11778