Theorem difrab2 30746

Description: Difference of two restricted class abstractions. Compare with difrab 4239. (Contributed by Thierry Arnoux, 3-Jan-2022.)

Assertion

Ref	Expression
difrab2	⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

Proof of Theorem difrab2

Step	Hyp	Ref	Expression
1		nfrab1 3310	. . 3 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜑}
2		nfrab1 3310	. . 3 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐵 ∣ 𝜑}
3	1, 2	nfdif 4056	. 2 ⊢ Ⅎ𝑥({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑})
4		nfrab1 3310	. 2 ⊢ Ⅎ𝑥{𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}
5		eldif 3893	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
6	5	anbi1i 623	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑))
7		andi 1004	. . . . . . 7 ⊢ ((𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ∨ (𝜑 ∧ ¬ 𝜑)))
8		pm3.24 402	. . . . . . . 8 ⊢ ¬ (𝜑 ∧ ¬ 𝜑)
9	8	biorfi 935	. . . . . . 7 ⊢ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ∨ (𝜑 ∧ ¬ 𝜑)))
10		ancom 460	. . . . . . 7 ⊢ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ↔ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑))
11	7, 9, 10	3bitr2i 298	. . . . . 6 ⊢ ((𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑))
12	11	anbi2i 622	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))) ↔ (𝑥 ∈ 𝐴 ∧ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑)))
13		anass 468	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))))
14		anass 468	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑)))
15	12, 13, 14	3bitr4i 302	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑))
16	6, 15	bitr4i 277	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
17		rabid 3304	. . 3 ⊢ (𝑥 ∈ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑} ↔ (𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑))
18		eldif 3893	. . . 4 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ ¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑}))
19		rabid 3304	. . . . 5 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
20		ianor 978	. . . . . 6 ⊢ (¬ (𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))
21		rabid 3304	. . . . . 6 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ (𝑥 ∈ 𝐵 ∧ 𝜑))
22	20, 21	xchnxbir 332	. . . . 5 ⊢ (¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))
23	19, 22	anbi12i 626	. . . 4 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ ¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
24	18, 23	bitri 274	. . 3 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
25	16, 17, 24	3bitr4ri 303	. 2 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ 𝑥 ∈ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑})
26	3, 4, 25	eqri 3937	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

Syntax hints:  ¬ wn 3   ∧ wa 395   ∨ wo 843   = wceq 1539   ∈ wcel 2108  {crab 3067   ∖ cdif 3880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-nf 1788  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-rab 3072  df-v 3424  df-dif 3886

This theorem is referenced by:  reprdifc  32507