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Theorem difininv 29872
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 ssdif0 4142 . . . . . 6 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
2 indif1 4072 . . . . . . 7 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
32eqeq1i 2804 . . . . . 6 (((𝐴𝐶) ∩ 𝐵) = ∅ ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
41, 3bitr4i 270 . . . . 5 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐶) ∩ 𝐵) = ∅)
54biimpri 220 . . . 4 (((𝐴𝐶) ∩ 𝐵) = ∅ → (𝐴𝐵) ⊆ 𝐶)
65adantr 473 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐶)
7 inss2 4029 . . . 4 (𝐴𝐵) ⊆ 𝐵
87a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐵)
96, 8ssind 4032 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ (𝐶𝐵))
10 ssdif0 4142 . . . . . 6 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
11 indif1 4072 . . . . . . 7 ((𝐶𝐴) ∩ 𝐵) = ((𝐶𝐵) ∖ 𝐴)
1211eqeq1i 2804 . . . . . 6 (((𝐶𝐴) ∩ 𝐵) = ∅ ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
1310, 12bitr4i 270 . . . . 5 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐴) ∩ 𝐵) = ∅)
1413biimpri 220 . . . 4 (((𝐶𝐴) ∩ 𝐵) = ∅ → (𝐶𝐵) ⊆ 𝐴)
1514adantl 474 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐴)
16 inss2 4029 . . . 4 (𝐶𝐵) ⊆ 𝐵
1716a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐵)
1815, 17ssind 4032 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ (𝐴𝐵))
199, 18eqssd 3815 1 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) = (𝐶𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 385   = wceq 1653  cdif 3766  cin 3768  wss 3769  c0 4115
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-ral 3094  df-rab 3098  df-v 3387  df-dif 3772  df-in 3776  df-ss 3783  df-nul 4116
This theorem is referenced by:  chtvalz  31227
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