Theorem csbab 4406

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 2709	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
2		sbsbc 3760	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
3	1, 2	bitri 275	. . 3 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
4		sbccom 3837	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
5		df-clab 2709	. . . . . 6 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
6		sbsbc 3760	. . . . . 6 ⊢ ([𝑧 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]𝜑)
7	5, 6	bitri 275	. . . . 5 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
8	7	sbcbii 3813	. . . 4 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
9	4, 8	bitr4i 278	. . 3 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑})
10		sbcel2 4384	. . 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 2727	1 ⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}

Syntax hints:   = wceq 1540  [wsb 2065   ∈ wcel 2109  {cab 2708  [wsbc 3756  ⦋csb 3865

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-nfc 2879  df-v 3452  df-sbc 3757  df-csb 3866  df-dif 3920  df-nul 4300

This theorem is referenced by:  csbsng  4675  csbuni  4903  csbxp  5741  csbdm  5864  csbfrecsg  8266  csbwrdg  14516  abfmpeld  32585  abfmpel  32586  csboprabg  37325  csbfinxpg  37383  csbingVD  44880  csbsngVD  44889  csbxpgVD  44890  csbrngVD  44892  csbunigVD  44894  csbfv12gALTVD  44895