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Theorem difin2 3226
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 2993 . . . . 5 (𝐴𝐶 → (𝑥𝐴𝑥𝐶))
21pm4.71d 385 . . . 4 (𝐴𝐶 → (𝑥𝐴 ↔ (𝑥𝐴𝑥𝐶)))
32anbi1d 452 . . 3 (𝐴𝐶 → ((𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵)))
4 eldif 2982 . . 3 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
5 elin 3155 . . . 4 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴))
6 eldif 2982 . . . . 5 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
76anbi1i 445 . . . 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 2079 1 (𝐴𝐶 → (𝐴𝐵) = ((𝐶𝐵) ∩ 𝐴))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102   = wceq 1284  wcel 1433  cdif 2970  cin 2972  wss 2973
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 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063
This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-dif 2975  df-in 2979  df-ss 2986
This theorem is referenced by: (None)
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