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Theorem difininv 29354
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 3942 . . . . . 6 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
2 indif1 3871 . . . . . . 7 ((𝐴𝐶) ∩ 𝐵) = ((𝐴𝐵) ∖ 𝐶)
32eqeq1i 2627 . . . . . 6 (((𝐴𝐶) ∩ 𝐵) = ∅ ↔ ((𝐴𝐵) ∖ 𝐶) = ∅)
41, 3bitr4i 267 . . . . 5 ((𝐴𝐵) ⊆ 𝐶 ↔ ((𝐴𝐶) ∩ 𝐵) = ∅)
54biimpri 218 . . . 4 (((𝐴𝐶) ∩ 𝐵) = ∅ → (𝐴𝐵) ⊆ 𝐶)
65adantr 481 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐶)
7 inss2 3834 . . . 4 (𝐴𝐵) ⊆ 𝐵
87a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ 𝐵)
96, 8ssind 3837 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) ⊆ (𝐶𝐵))
10 ssdif0 3942 . . . . . 6 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
11 indif1 3871 . . . . . . 7 ((𝐶𝐴) ∩ 𝐵) = ((𝐶𝐵) ∖ 𝐴)
1211eqeq1i 2627 . . . . . 6 (((𝐶𝐴) ∩ 𝐵) = ∅ ↔ ((𝐶𝐵) ∖ 𝐴) = ∅)
1310, 12bitr4i 267 . . . . 5 ((𝐶𝐵) ⊆ 𝐴 ↔ ((𝐶𝐴) ∩ 𝐵) = ∅)
1413biimpri 218 . . . 4 (((𝐶𝐴) ∩ 𝐵) = ∅ → (𝐶𝐵) ⊆ 𝐴)
1514adantl 482 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐴)
16 inss2 3834 . . . 4 (𝐶𝐵) ⊆ 𝐵
1716a1i 11 . . 3 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ 𝐵)
1815, 17ssind 3837 . 2 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐶𝐵) ⊆ (𝐴𝐵))
199, 18eqssd 3620 1 ((((𝐴𝐶) ∩ 𝐵) = ∅ ∧ ((𝐶𝐴) ∩ 𝐵) = ∅) → (𝐴𝐵) = (𝐶𝐵))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 384   = wceq 1483  cdif 3571  cin 3573  wss 3574  c0 3915
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ral 2917  df-rab 2921  df-v 3202  df-dif 3577  df-in 3581  df-ss 3588  df-nul 3916
This theorem is referenced by:  chtvalz  30707
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