Theorem csbabg 2901

Description: Move substitution into a class abstraction. (Contributed by NM, 13-Dec-2005.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)

Assertion

Ref	Expression
csbabg	⊢ (A ∈ 𝑉 → ⦋A / x⦌{y ∣ φ} = {y ∣ [A / x]φ})

Distinct variable groups:   y,A   x,y

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

Proof of Theorem csbabg

Dummy variable z is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbccom 2827	. . . 4 ⊢ ([z / y][A / x]φ ↔ [A / x][z / y]φ)
2		df-clab 2024	. . . . 5 ⊢ (z ∈ {y ∣ [A / x]φ} ↔ [z / y][A / x]φ)
3		sbsbc 2762	. . . . 5 ⊢ ([z / y][A / x]φ ↔ [z / y][A / x]φ)
4	2, 3	bitri 173	. . . 4 ⊢ (z ∈ {y ∣ [A / x]φ} ↔ [z / y][A / x]φ)
5		df-clab 2024	. . . . . 6 ⊢ (z ∈ {y ∣ φ} ↔ [z / y]φ)
6		sbsbc 2762	. . . . . 6 ⊢ ([z / y]φ ↔ [z / y]φ)
7	5, 6	bitri 173	. . . . 5 ⊢ (z ∈ {y ∣ φ} ↔ [z / y]φ)
8	7	sbcbii 2812	. . . 4 ⊢ ([A / x]z ∈ {y ∣ φ} ↔ [A / x][z / y]φ)
9	1, 4, 8	3bitr4i 201	. . 3 ⊢ (z ∈ {y ∣ [A / x]φ} ↔ [A / x]z ∈ {y ∣ φ})
10		sbcel2g 2865	. . 3 ⊢ (A ∈ 𝑉 → ([A / x]z ∈ {y ∣ φ} ↔ z ∈ ⦋A / x⦌{y ∣ φ}))
11	9, 10	syl5rbb 182	. 2 ⊢ (A ∈ 𝑉 → (z ∈ ⦋A / x⦌{y ∣ φ} ↔ z ∈ {y ∣ [A / x]φ}))
12	11	eqrdv 2035	1 ⊢ (A ∈ 𝑉 → ⦋A / x⦌{y ∣ φ} = {y ∣ [A / x]φ})

Syntax hints:   → wi 4   = wceq 1242   ∈ wcel 1390  [wsb 1642  {cab 2023  [wsbc 2758  ⦋csb 2846

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-sbc 2759  df-csb 2847

This theorem is referenced by:  csbsng  3422  csbunig  3579  csbxpg  4364  csbdmg  4472  csbrng  4725