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Theorem difin2 4216
Description: Represent a class 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 3908 . . . . 5 (𝐴𝐶 → (𝑥𝐴𝑥𝐶))
21pm4.71d 565 . . . 4 (𝐴𝐶 → (𝑥𝐴 ↔ (𝑥𝐴𝑥𝐶)))
32anbi1d 632 . . 3 (𝐴𝐶 → ((𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵)))
4 eldif 3891 . . 3 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
5 elin 3897 . . . 4 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴))
6 eldif 3891 . . . . 5 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
76anbi1i 626 . . . 4 ((𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴) ↔ ((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴))
8 ancom 464 . . . . 5 (((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴) ↔ (𝑥𝐴 ∧ (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
9 anass 472 . . . . 5 (((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵) ↔ (𝑥𝐴 ∧ (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
108, 9bitr4i 281 . . . 4 (((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵))
115, 7, 103bitri 300 . . 3 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵))
123, 4, 113bitr4g 317 . 2 (𝐴𝐶 → (𝑥 ∈ (𝐴𝐵) ↔ 𝑥 ∈ ((𝐶𝐵) ∩ 𝐴)))
1312eqrdv 2796 1 (𝐴𝐶 → (𝐴𝐵) = ((𝐶𝐵) ∩ 𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 399   = wceq 1538  wcel 2111  cdif 3878  cin 3880  wss 3881
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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770
This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-v 3443  df-dif 3884  df-in 3888  df-ss 3898
This theorem is referenced by:  gsumdifsnd  19074  issubdrg  19553  restcld  21777  limcnlp  24481  symgcom2  30778  difelsiga  31502  sigapildsyslem  31530  ldgenpisyslem1  31532  difelcarsg2  31681  ballotlemfp1  31859  asindmre  35140  caragendifcl  43153  gsumdifsndf  44441
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