Theorem difrab2 32479

Description: Difference of two restricted class abstractions. Compare with difrab 4293. (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 3436	. . 3 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜑}
2		nfrab1 3436	. . 3 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐵 ∣ 𝜑}
3	1, 2	nfdif 4104	. 2 ⊢ Ⅎ𝑥({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑})
4		nfrab1 3436	. 2 ⊢ Ⅎ𝑥{𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}
5		eldif 3936	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
6	5	anbi1i 624	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑))
7		andi 1009	. . . . . . 7 ⊢ ((𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ∨ (𝜑 ∧ ¬ 𝜑)))
8		pm3.24 402	. . . . . . . 8 ⊢ ¬ (𝜑 ∧ ¬ 𝜑)
9	8	biorfri 939	. . . . . . 7 ⊢ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ∨ (𝜑 ∧ ¬ 𝜑)))
10		ancom 460	. . . . . . 7 ⊢ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ↔ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑))
11	7, 9, 10	3bitr2i 299	. . . . . 6 ⊢ ((𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑))
12	11	anbi2i 623	. . . . 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 303	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑))
16	6, 15	bitr4i 278	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
17		rabid 3437	. . 3 ⊢ (𝑥 ∈ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑} ↔ (𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑))
18		eldif 3936	. . . 4 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ ¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑}))
19		rabid 3437	. . . . 5 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
20		ianor 983	. . . . . 6 ⊢ (¬ (𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))
21		rabid 3437	. . . . . 6 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ (𝑥 ∈ 𝐵 ∧ 𝜑))
22	20, 21	xchnxbir 333	. . . . 5 ⊢ (¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))
23	19, 22	anbi12i 628	. . . 4 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ ¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
24	18, 23	bitri 275	. . 3 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
25	16, 17, 24	3bitr4ri 304	. 2 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ 𝑥 ∈ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑})
26	3, 4, 25	eqri 3979	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

Syntax hints:  ¬ wn 3   ∧ wa 395   ∨ wo 847   = wceq 1540   ∈ wcel 2108  {crab 3415   ∖ cdif 3923

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1543  df-ex 1780  df-nf 1784  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-nfc 2885  df-rab 3416  df-v 3461  df-dif 3929

This theorem is referenced by:  reprdifc  34659