Theorem csbab 4371

Description: Move substitution into a class abstraction. (Contributed by NM, 13-Dec-2005.) (Revised by NM, 19-Aug-2018.)

Assertion

Ref	Expression
csbab	⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}

Distinct variable groups:   𝑦,𝐴   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝐴(𝑥)

Proof of Theorem csbab

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-clab 2716	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
2		sbsbc 3720	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
3	1, 2	bitri 274	. . 3 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
4		sbccom 3804	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
5		df-clab 2716	. . . . . 6 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
6		sbsbc 3720	. . . . . 6 ⊢ ([𝑧 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]𝜑)
7	5, 6	bitri 274	. . . . 5 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
8	7	sbcbii 3776	. . . 4 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
9	4, 8	bitr4i 277	. . 3 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑})
10		sbcel2 4349	. . 3 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ 𝑧 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑})
11	3, 9, 10	3bitrri 298	. 2 ⊢ (𝑧 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} ↔ 𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑})
12	11	eqriv 2735	1 ⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}

Syntax hints:   = wceq 1539  [wsb 2067   ∈ wcel 2106  {cab 2715  [wsbc 3716  ⦋csb 3832

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-nfc 2889  df-v 3434  df-sbc 3717  df-csb 3833  df-dif 3890  df-nul 4257

This theorem is referenced by:  csbsng  4644  csbuni  4870  csbxp  5686  csbdm  5806  csbfrecsg  8100  csbwrdg  14247  abfmpeld  30991  abfmpel  30992  csboprabg  35501  csbfinxpg  35559  csbingVD  42504  csbsngVD  42513  csbxpgVD  42514  csbrngVD  42516  csbunigVD  42518  csbfv12gALTVD  42519