Theorem sbcel1g 3114

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 3110	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ∈ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ∈ ⦋𝐴 / 𝑥⦌𝐶))
2		csbconstg 3109	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝐶 = 𝐶)
3	2	eleq2d 2276	. 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 2177  [wsbc 3000  ⦋csb 3095

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-sbc 3001  df-csb 3096

This theorem is referenced by:  rspcsbela  3155  fprodcllemf  11974