Theorem csbab 4397

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 2744	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
2		sbsbc 3751	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
3	1, 2	bitri 278	. . 3 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
4		sbccom 3827	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
5		df-clab 2744	. . . . . 6 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
6		sbsbc 3751	. . . . . 6 ⊢ ([𝑧 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]𝜑)
7	5, 6	bitri 278	. . . . 5 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
8	7	sbcbii 3803	. . . 4 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
9	4, 8	bitr4i 281	. . 3 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑})
10		sbcel2 4375	. . 3 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ 𝑧 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑})
11	3, 9, 10	3bitrri 301	. 2 ⊢ (𝑧 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} ↔ 𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑})
12	11	eqriv 2762	1 ⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}

Syntax hints:   = wceq 1563  [wsb 2093   ∈ wcel 2145  {cab 2743  [wsbc 3747  ⦋csb 3855

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2737

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1566  df-fal 1576  df-ex 1803  df-nf 1807  df-sb 2094  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-v 3459  df-sbc 3748  df-csb 3856  df-dif 3910  df-nul 4289

This theorem is referenced by:  csbsng  4670  csbuni  4899  csbxp  5753  csbdm  5878  csbfrecsg  8269  csbwrdg  14571  abfmpeld  32911  abfmpel  32912  csboprabg  37836  csbfinxpg  37894  csbingVD  45457  csbsngVD  45466  csbxpgVD  45467  csbrngVD  45469  csbunigVD  45471  csbfv12gALTVD  45472