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Theorem bj-inrab 32115
 Description: Generalization of inrab 3858. (Contributed by BJ, 21-Apr-2019.)
Assertion
Ref Expression
bj-inrab ({𝑥𝐴𝜑} ∩ {𝑥𝐵𝜓}) = {𝑥 ∈ (𝐴𝐵) ∣ (𝜑𝜓)}

Proof of Theorem bj-inrab
StepHypRef Expression
1 an4 861 . . . 4 (((𝑥𝐴𝜑) ∧ (𝑥𝐵𝜓)) ↔ ((𝑥𝐴𝑥𝐵) ∧ (𝜑𝜓)))
2 elin 3758 . . . . 5 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴𝑥𝐵))
32anbi1i 727 . . . 4 ((𝑥 ∈ (𝐴𝐵) ∧ (𝜑𝜓)) ↔ ((𝑥𝐴𝑥𝐵) ∧ (𝜑𝜓)))
41, 3bitr4i 266 . . 3 (((𝑥𝐴𝜑) ∧ (𝑥𝐵𝜓)) ↔ (𝑥 ∈ (𝐴𝐵) ∧ (𝜑𝜓)))
54abbii 2726 . 2 {𝑥 ∣ ((𝑥𝐴𝜑) ∧ (𝑥𝐵𝜓))} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ (𝜑𝜓))}
6 df-rab 2905 . . . 4 {𝑥𝐴𝜑} = {𝑥 ∣ (𝑥𝐴𝜑)}
7 df-rab 2905 . . . 4 {𝑥𝐵𝜓} = {𝑥 ∣ (𝑥𝐵𝜓)}
86, 7ineq12i 3774 . . 3 ({𝑥𝐴𝜑} ∩ {𝑥𝐵𝜓}) = ({𝑥 ∣ (𝑥𝐴𝜑)} ∩ {𝑥 ∣ (𝑥𝐵𝜓)})
9 inab 3854 . . 3 ({𝑥 ∣ (𝑥𝐴𝜑)} ∩ {𝑥 ∣ (𝑥𝐵𝜓)}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∧ (𝑥𝐵𝜓))}
108, 9eqtri 2632 . 2 ({𝑥𝐴𝜑} ∩ {𝑥𝐵𝜓}) = {𝑥 ∣ ((𝑥𝐴𝜑) ∧ (𝑥𝐵𝜓))}
11 df-rab 2905 . 2 {𝑥 ∈ (𝐴𝐵) ∣ (𝜑𝜓)} = {𝑥 ∣ (𝑥 ∈ (𝐴𝐵) ∧ (𝜑𝜓))}
125, 10, 113eqtr4i 2642 1 ({𝑥𝐴𝜑} ∩ {𝑥𝐵𝜓}) = {𝑥 ∈ (𝐴𝐵) ∣ (𝜑𝜓)}
 Colors of variables: wff setvar class Syntax hints:   ∧ wa 383   = wceq 1475   ∈ wcel 1977  {cab 2596  {crab 2900   ∩ cin 3539 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-rab 2905  df-v 3175  df-in 3547 This theorem is referenced by:  bj-inrab2  32116
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