Theorem csbab 4151

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 2747	. . . 4 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
2		sbsbc 3580	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
3	1, 2	bitri 264	. . 3 ⊢ (𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑} ↔ [𝑧 / 𝑦][𝐴 / 𝑥]𝜑)
4		sbccom 3650	. . . 4 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
5		df-clab 2747	. . . . . 6 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
6		sbsbc 3580	. . . . . 6 ⊢ ([𝑧 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]𝜑)
7	5, 6	bitri 264	. . . . 5 ⊢ (𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝑧 / 𝑦]𝜑)
8	7	sbcbii 3632	. . . 4 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ [𝐴 / 𝑥][𝑧 / 𝑦]𝜑)
9	4, 8	bitr4i 267	. . 3 ⊢ ([𝑧 / 𝑦][𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑})
10		sbcel2 4132	. . 3 ⊢ ([𝐴 / 𝑥]𝑧 ∈ {𝑦 ∣ 𝜑} ↔ 𝑧 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑})
11	3, 9, 10	3bitrri 287	. 2 ⊢ (𝑧 ∈ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} ↔ 𝑧 ∈ {𝑦 ∣ [𝐴 / 𝑥]𝜑})
12	11	eqriv 2757	1 ⊢ ⦋𝐴 / 𝑥⦌{𝑦 ∣ 𝜑} = {𝑦 ∣ [𝐴 / 𝑥]𝜑}

Syntax hints:   = wceq 1632  [wsb 2046   ∈ wcel 2139  {cab 2746  [wsbc 3576  ⦋csb 3674

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-tru 1635  df-fal 1638  df-ex 1854  df-nf 1859  df-sb 2047  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-v 3342  df-sbc 3577  df-csb 3675  df-dif 3718  df-nul 4059

This theorem is referenced by:  csbsng  4387  csbuni  4618  csbxp  5357  csbdm  5473  csbwrdg  13520  abfmpeld  29763  abfmpel  29764  csbwrecsg  33484  csboprabg  33487  csbfinxpg  33536  csbxpgOLD  39553  csbrngOLD  39556  csbfv12gALTVD  39634