Theorem difrab2 33142

Description: Difference of two restricted class abstractions. Compare with difrab 4263. (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 3431	. . 3 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐴 ∣ 𝜑}
2		nfrab1 3431	. . 3 ⊢ Ⅎ𝑥{𝑥 ∈ 𝐵 ∣ 𝜑}
3	1, 2	nfdif 4066	. 2 ⊢ Ⅎ𝑥({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑})
4		nfrab1 3431	. 2 ⊢ Ⅎ𝑥{𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}
5		eldif 4049	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))
6	5	anbi1i 636	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑))
7		andi 1025	. . . . . . 7 ⊢ ((𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ∨ (𝜑 ∧ ¬ 𝜑)))
8		pm3.24 408	. . . . . . . 8 ⊢ ¬ (𝜑 ∧ ¬ 𝜑)
9	8	biorfri 953	. . . . . . 7 ⊢ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ∨ (𝜑 ∧ ¬ 𝜑)))
10		ancom 466	. . . . . . 7 ⊢ ((𝜑 ∧ ¬ 𝑥 ∈ 𝐵) ↔ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑))
11	7, 9, 10	3bitr2i 302	. . . . . 6 ⊢ ((𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑))
12	11	anbi2i 635	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))) ↔ (𝑥 ∈ 𝐴 ∧ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑)))
13		anass 474	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ (𝜑 ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))))
14		anass 474	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (¬ 𝑥 ∈ 𝐵 ∧ 𝜑)))
15	12, 13, 14	3bitr4i 306	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)) ↔ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ∧ 𝜑))
16	6, 15	bitr4i 281	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
17		rabid 3432	. . 3 ⊢ (𝑥 ∈ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑} ↔ (𝑥 ∈ (𝐴 ∖ 𝐵) ∧ 𝜑))
18		eldif 4049	. . . 4 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ ¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑}))
19		rabid 3432	. . . . 5 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ↔ (𝑥 ∈ 𝐴 ∧ 𝜑))
20		ianor 997	. . . . . 6 ⊢ (¬ (𝑥 ∈ 𝐵 ∧ 𝜑) ↔ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))
21		rabid 3432	. . . . . 6 ⊢ (𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ (𝑥 ∈ 𝐵 ∧ 𝜑))
22	20, 21	xchnxbir 336	. . . . 5 ⊢ (¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑} ↔ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑))
23	19, 22	anbi12i 640	. . . 4 ⊢ ((𝑥 ∈ {𝑥 ∈ 𝐴 ∣ 𝜑} ∧ ¬ 𝑥 ∈ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
24	18, 23	bitri 278	. . 3 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ ((𝑥 ∈ 𝐴 ∧ 𝜑) ∧ (¬ 𝑥 ∈ 𝐵 ∨ ¬ 𝜑)))
25	16, 17, 24	3bitr4ri 307	. 2 ⊢ (𝑥 ∈ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) ↔ 𝑥 ∈ {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑})
26	3, 4, 25	eqri 3928	1 ⊢ ({𝑥 ∈ 𝐴 ∣ 𝜑} ∖ {𝑥 ∈ 𝐵 ∣ 𝜑}) = {𝑥 ∈ (𝐴 ∖ 𝐵) ∣ 𝜑}

This proof depends on syntax axioms:  ¬ wn 3   ∧ wa 401   ∨ wo 861   = wceq 1569   ∈ wcel 2144  {crab 3412   ∖ cdif 4045

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 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-rab 3413  df-v 3452  df-dif 4047

This theorem is used by:  reprdifc  35303