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Theorem difin2 3259
Description: Represent a set difference as an intersection with a larger difference. (Contributed by Jeff Madsen, 2-Sep-2009.)
Assertion
Ref Expression
difin2 (𝐴𝐶 → (𝐴𝐵) = ((𝐶𝐵) ∩ 𝐴))

Proof of Theorem difin2
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 ssel 3017 . . . . 5 (𝐴𝐶 → (𝑥𝐴𝑥𝐶))
21pm4.71d 385 . . . 4 (𝐴𝐶 → (𝑥𝐴 ↔ (𝑥𝐴𝑥𝐶)))
32anbi1d 453 . . 3 (𝐴𝐶 → ((𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵)))
4 eldif 3006 . . 3 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
5 elin 3181 . . . 4 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴))
6 eldif 3006 . . . . 5 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
76anbi1i 446 . . . 4 ((𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴) ↔ ((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴))
8 ancom 262 . . . . 5 (((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴) ↔ (𝑥𝐴 ∧ (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
9 anass 393 . . . . 5 (((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵) ↔ (𝑥𝐴 ∧ (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
108, 9bitr4i 185 . . . 4 (((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵))
115, 7, 103bitri 204 . . 3 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵))
123, 4, 113bitr4g 221 . 2 (𝐴𝐶 → (𝑥 ∈ (𝐴𝐵) ↔ 𝑥 ∈ ((𝐶𝐵) ∩ 𝐴)))
1312eqrdv 2086 1 (𝐴𝐶 → (𝐴𝐵) = ((𝐶𝐵) ∩ 𝐴))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102   = wceq 1289  wcel 1438  cdif 2994  cin 2996  wss 2997
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 579  ax-in2 580  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-v 2621  df-dif 2999  df-in 3003  df-ss 3010
This theorem is referenced by: (None)
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