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Theorem difininv 30266
 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 4223 . . . . . 6 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
21eqeq1i 2826 . . . . 5 (((𝐴𝐶) ∩ 𝐵) = ∅ ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
3 ssdif0 4296 . . . . 5 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
42, 3sylbb2 241 . . . 4 (((𝐴𝐶) ∩ 𝐵) = ∅ → (𝐴𝐵) ⊆ 𝐶)
54adantr 484 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐶)
6 inss2 4181 . . . 4 (𝐴𝐵) ⊆ 𝐵
76a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐵)
85, 7ssind 4184 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ (𝐶𝐵))
9 indif1 4223 . . . . . 6 ((𝐶𝐴) ∩ 𝐵) = ((𝐶𝐵) ∖ 𝐴)
109eqeq1i 2826 . . . . 5 (((𝐶𝐴) ∩ 𝐵) = ∅ ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
11 ssdif0 4296 . . . . 5 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
1210, 11sylbb2 241 . . . 4 (((𝐶𝐴) ∩ 𝐵) = ∅ → (𝐶𝐵) ⊆ 𝐴)
1312adantl 485 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐴)
14 inss2 4181 . . . 4 (𝐶𝐵) ⊆ 𝐵
1514a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐵)
1613, 15ssind 4184 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ (𝐴𝐵))
178, 16eqssd 3960 1 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) = (𝐶𝐵))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ wa 399   = wceq 1538   ∖ cdif 3907   ∩ cin 3909   ⊆ wss 3910  ∅c0 4266 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-11 2162  ax-12 2178  ax-ext 2793 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2071  df-clab 2800  df-cleq 2814  df-clel 2892  df-nfc 2960  df-rab 3135  df-v 3473  df-dif 3913  df-in 3917  df-ss 3927  df-nul 4267 This theorem is referenced by:  chtvalz  31908
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