Theorem difininv 32995

Description: Condition for the intersections of two sets with a given set to be equal. (Contributed by Thierry Arnoux, 28-Dec-2021.)

Assertion

Ref	Expression
difininv	⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐴 ∩ 𝐵) = (𝐶 ∩ 𝐵))

Proof of Theorem difininv

Step	Hyp	Ref	Expression
1		indif1 4228	. . . . . 6 ⊢ ((𝐴 ∖ 𝐶) ∩ 𝐵) = ((𝐴 ∩ 𝐵) ∖ 𝐶)
2	1	eqeq1i 2765	. . . . 5 ⊢ (((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ↔ ((𝐴 ∩ 𝐵) ∖ 𝐶) = ∅)
3		ssdif0 4314	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ⊆ 𝐶 ↔ ((𝐴 ∩ 𝐵) ∖ 𝐶) = ∅)
4	2, 3	sylbb2 241	. . . 4 ⊢ (((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ → (𝐴 ∩ 𝐵) ⊆ 𝐶)
5	4	adantr 486	. . 3 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐴 ∩ 𝐵) ⊆ 𝐶)
6		inss2 4183	. . . 4 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐵
7	6	a1i 11	. . 3 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐴 ∩ 𝐵) ⊆ 𝐵)
8	5, 7	ssind 4186	. 2 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐴 ∩ 𝐵) ⊆ (𝐶 ∩ 𝐵))
9		indif1 4228	. . . . . 6 ⊢ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ((𝐶 ∩ 𝐵) ∖ 𝐴)
10	9	eqeq1i 2765	. . . . 5 ⊢ (((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅ ↔ ((𝐶 ∩ 𝐵) ∖ 𝐴) = ∅)
11		ssdif0 4314	. . . . 5 ⊢ ((𝐶 ∩ 𝐵) ⊆ 𝐴 ↔ ((𝐶 ∩ 𝐵) ∖ 𝐴) = ∅)
12	10, 11	sylbb2 241	. . . 4 ⊢ (((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅ → (𝐶 ∩ 𝐵) ⊆ 𝐴)
13	12	adantl 487	. . 3 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐶 ∩ 𝐵) ⊆ 𝐴)
14		inss2 4183	. . . 4 ⊢ (𝐶 ∩ 𝐵) ⊆ 𝐵
15	14	a1i 11	. . 3 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐶 ∩ 𝐵) ⊆ 𝐵)
16	13, 15	ssind 4186	. 2 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐶 ∩ 𝐵) ⊆ (𝐴 ∩ 𝐵))
17	8, 16	eqssd 3948	1 ⊢ ((((𝐴 ∖ 𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶 ∖ 𝐴) ∩ 𝐵) = ∅) → (𝐴 ∩ 𝐵) = (𝐶 ∩ 𝐵))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∖ cdif 3896   ∩ cin 3898   ⊆ wss 3899  ∅c0 4279

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-dif 3902  df-in 3906  df-ss 3916  df-nul 4280

This theorem is used by:  chtvalz  35140