Theorem csbab 4393

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 2708	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
2		sbsbc 3748	. . . 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 3825	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
5		df-clab 2708	. . . . . 6 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
6		sbsbc 3748	. . . . . 6 ⊢ ([𝑧 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]𝜑)
7	5, 6	bitri 275	. . . . 5 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
8	7	sbcbii 3801	. . . 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 4371	. . 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 2726	1 ⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}

Syntax hints:   = wceq 1540  [wsb 2065   ∈ wcel 2109  {cab 2707  [wsbc 3744  ⦋csb 3853

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 2701

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 2708  df-cleq 2721  df-clel 2803  df-nfc 2878  df-v 3440  df-sbc 3745  df-csb 3854  df-dif 3908  df-nul 4287

This theorem is referenced by:  csbsng  4662  csbuni  4890  csbxp  5723  csbdm  5844  csbfrecsg  8224  csbwrdg  14469  abfmpeld  32611  abfmpel  32612  csboprabg  37303  csbfinxpg  37361  csbingVD  44857  csbsngVD  44866  csbxpgVD  44867  csbrngVD  44869  csbunigVD  44871  csbfv12gALTVD  44872