Theorem inssdif0OLD 4322

Description: Obsolete version of inssdif0 4321 as of 18-Jul-2026. (Contributed by NM, 27-Oct-1996.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof shortened by Wolf Lammen, 30-Sep-2014.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
inssdif0OLD	⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅)

Proof of Theorem inssdif0OLD

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elin 4093	. . . . . 6 ⊢ (𝑥 ∈ (𝐴 ∩ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵))
2	1	imbi1i 352	. . . . 5 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐶) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐶))
3		iman 407	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) → 𝑥 ∈ 𝐶) ↔ ¬ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ ¬ 𝑥 ∈ 𝐶))
4	2, 3	bitri 278	. . . 4 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐶) ↔ ¬ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ ¬ 𝑥 ∈ 𝐶))
5		eldif 4049	. . . . . 6 ⊢ (𝑥 ∈ (𝐵 ∖ 𝐶) ↔ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐶))
6	5	anbi2i 635	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∖ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐶)))
7		elin 4093	. . . . 5 ⊢ (𝑥 ∈ (𝐴 ∩ (𝐵 ∖ 𝐶)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ (𝐵 ∖ 𝐶)))
8		anass 474	. . . . 5 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ ¬ 𝑥 ∈ 𝐶) ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐶)))
9	6, 7, 8	3bitr4ri 307	. . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ∧ ¬ 𝑥 ∈ 𝐶) ↔ 𝑥 ∈ (𝐴 ∩ (𝐵 ∖ 𝐶)))
10	4, 9	xchbinx 337	. . 3 ⊢ ((𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐶) ↔ ¬ 𝑥 ∈ (𝐴 ∩ (𝐵 ∖ 𝐶)))
11	10	albii 1851	. 2 ⊢ (∀𝑥(𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐶) ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵 ∖ 𝐶)))
12		df-ss 3896	. 2 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ (𝐴 ∩ 𝐵) → 𝑥 ∈ 𝐶))
13		eq0 4296	. 2 ⊢ ((𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅ ↔ ∀𝑥 ¬ 𝑥 ∈ (𝐴 ∩ (𝐵 ∖ 𝐶)))
14	11, 12, 13	3bitr4i 306	1 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ (𝐴 ∩ (𝐵 ∖ 𝐶)) = ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567   = wceq 1569   ∈ wcel 2144   ⊆ wss 3895   ∖ cdif 4045   ∩ cin 4089  ∅c0 4278

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-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-nul 4279

This theorem is used by: (None)