Theorem sbcel1g 3068

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 3064	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3063	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eleq2d 2240	. 2 ⊢ (𝐴 ∈ 𝑉 → (⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ 𝐶))
4	1, 3	bitrd 187	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ 𝐶))

Syntax hints:   → wi 4   ↔ wb 104   ∈ wcel 2141  [wsbc 2955  ⦋csb 3049

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-sbc 2956  df-csb 3050

This theorem is referenced by:  rspcsbela  3108  fprodcllemf  11576