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Theorem difininv 32022
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 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) = (𝐶𝐵))

Proof of Theorem difininv
StepHypRef Expression
1 indif1 4270 . . . . . 6 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
21eqeq1i 2735 . . . . 5 (((𝐴𝐶) ∩ 𝐵) = ∅ ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
3 ssdif0 4362 . . . . 5 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
42, 3sylbb2 237 . . . 4 (((𝐴𝐶) ∩ 𝐵) = ∅ → (𝐴𝐵) ⊆ 𝐶)
54adantr 479 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐶)
6 inss2 4228 . . . 4 (𝐴𝐵) ⊆ 𝐵
76a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐵)
85, 7ssind 4231 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ (𝐶𝐵))
9 indif1 4270 . . . . . 6 ((𝐶𝐴) ∩ 𝐵) = ((𝐶𝐵) ∖ 𝐴)
109eqeq1i 2735 . . . . 5 (((𝐶𝐴) ∩ 𝐵) = ∅ ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
11 ssdif0 4362 . . . . 5 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
1210, 11sylbb2 237 . . . 4 (((𝐶𝐴) ∩ 𝐵) = ∅ → (𝐶𝐵) ⊆ 𝐴)
1312adantl 480 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐴)
14 inss2 4228 . . . 4 (𝐶𝐵) ⊆ 𝐵
1514a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐵)
1613, 15ssind 4231 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ (𝐴𝐵))
178, 16eqssd 3998 1 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) = (𝐶𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 394   = wceq 1539  cdif 3944  cin 3946  wss 3947  c0 4321
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 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701
This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1542  df-fal 1552  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-rab 3431  df-v 3474  df-dif 3950  df-in 3954  df-ss 3964  df-nul 4322
This theorem is referenced by:  chtvalz  33939
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