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Theorem dfif6 4487
Description: An alternate definition of the conditional operator df-if 4485 as a simple class abstraction. (Contributed by Mario Carneiro, 8-Sep-2013.)
Assertion
Ref Expression
dfif6 if(𝜑, 𝐴, 𝐵) = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ ¬ 𝜑})
Distinct variable groups:   𝜑,𝑥   𝑥,𝐴   𝑥,𝐵

Proof of Theorem dfif6
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 eleq1w 2846 . . . 4 (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
21anbi1d 643 . . 3 (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑦 ∈ 𝐴 ∧ 𝜑)))
3 eleq1w 2846 . . . 4 (𝑥 = 𝑦 → (𝑥 ∈ 𝐵 ↔ 𝑦 ∈ 𝐵))
43anbi1d 643 . . 3 (𝑥 = 𝑦 → ((𝑥 ∈ 𝐵 ∧ ¬ 𝜑) ↔ (𝑦 ∈ 𝐵 ∧ ¬ 𝜑)))
52, 4unabw 4254 . 2 ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ ¬ 𝜑)}) = {𝑦 ∣ ((𝑦 ∈ 𝐴 ∧ 𝜑) ∨ (𝑦 ∈ 𝐵 ∧ ¬ 𝜑))}
6 df-rab 3416 . . 3 {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
7 df-rab 3416 . . 3 {𝑥 ∈ 𝐵 ∣ ¬ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ ¬ 𝜑)}
86, 7uneq12i 4167 . 2 ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ ¬ 𝜑}) = ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} ∪ {𝑥 ∣ (𝑥 ∈ 𝐵 ∧ ¬ 𝜑)})
9 df-if 4485 . 2 if(𝜑, 𝐴, 𝐵) = {𝑦 ∣ ((𝑦 ∈ 𝐴 ∧ 𝜑) ∨ (𝑦 ∈ 𝐵 ∧ ¬ 𝜑))}
105, 8, 93eqtr4ri 2797 1 if(𝜑, 𝐴, 𝐵) = ({𝑥 ∈ 𝐴 ∣ 𝜑} ∪ {𝑥 ∈ 𝐵 ∣ ¬ 𝜑})
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401   ∨ wo 861   = wceq 1569   ∈ wcel 2146  {cab 2741  {crab 3415   ∪ cun 4151  ifcif 4484
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3416  df-v 3455  df-un 4153  df-if 4485
This theorem is used by:  ifeq1  4488  ifeq2  4489  dfif3  4499
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