Theorem inab 3199

Description: Intersection of two class abstractions. (Contributed by NM, 29-Sep-2002.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
inab	⊢ ({x ∣ φ} ∩ {x ∣ ψ}) = {x ∣ (φ ∧ ψ)}

Proof of Theorem inab

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sban 1826	. . 3 ⊢ ([y / x](φ ∧ ψ) ↔ ([y / x]φ ∧ [y / x]ψ))
2		df-clab 2024	. . 3 ⊢ (y ∈ {x ∣ (φ ∧ ψ)} ↔ [y / x](φ ∧ ψ))
3		df-clab 2024	. . . 4 ⊢ (y ∈ {x ∣ φ} ↔ [y / x]φ)
4		df-clab 2024	. . . 4 ⊢ (y ∈ {x ∣ ψ} ↔ [y / x]ψ)
5	3, 4	anbi12i 433	. . 3 ⊢ ((y ∈ {x ∣ φ} ∧ y ∈ {x ∣ ψ}) ↔ ([y / x]φ ∧ [y / x]ψ))
6	1, 2, 5	3bitr4ri 202	. 2 ⊢ ((y ∈ {x ∣ φ} ∧ y ∈ {x ∣ ψ}) ↔ y ∈ {x ∣ (φ ∧ ψ)})
7	6	ineqri 3124	1 ⊢ ({x ∣ φ} ∩ {x ∣ ψ}) = {x ∣ (φ ∧ ψ)}

Syntax hints:   ∧ wa 97   = wceq 1242   ∈ wcel 1390  [wsb 1642  {cab 2023   ∩ cin 2910

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-bndl 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-in 2918

This theorem is referenced by:  inrab  3203  inrab2  3204  dfrab2  3206  dfrab3  3207  imainlem  4923  imain  4924