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Theorem difininv 32545
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 4288 . . . . . 6 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
21eqeq1i 2740 . . . . 5 (((𝐴𝐶) ∩ 𝐵) = ∅ ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
3 ssdif0 4372 . . . . 5 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
42, 3sylbb2 238 . . . 4 (((𝐴𝐶) ∩ 𝐵) = ∅ → (𝐴𝐵) ⊆ 𝐶)
54adantr 480 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐶)
6 inss2 4246 . . . 4 (𝐴𝐵) ⊆ 𝐵
76a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐵)
85, 7ssind 4249 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ (𝐶𝐵))
9 indif1 4288 . . . . . 6 ((𝐶𝐴) ∩ 𝐵) = ((𝐶𝐵) ∖ 𝐴)
109eqeq1i 2740 . . . . 5 (((𝐶𝐴) ∩ 𝐵) = ∅ ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
11 ssdif0 4372 . . . . 5 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
1210, 11sylbb2 238 . . . 4 (((𝐶𝐴) ∩ 𝐵) = ∅ → (𝐶𝐵) ⊆ 𝐴)
1312adantl 481 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐴)
14 inss2 4246 . . . 4 (𝐶𝐵) ⊆ 𝐵
1514a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐵)
1613, 15ssind 4249 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ (𝐴𝐵))
178, 16eqssd 4013 1 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) = (𝐶𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 395   = wceq 1537  cdif 3960  cin 3962  wss 3963  c0 4339
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-fal 1550  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-rab 3434  df-v 3480  df-dif 3966  df-in 3970  df-ss 3980  df-nul 4340
This theorem is referenced by:  chtvalz  34623
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