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Theorem difin2 4091
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 3792 . . . . 5 (𝐴𝐶 → (𝑥𝐴𝑥𝐶))
21pm4.71d 553 . . . 4 (𝐴𝐶 → (𝑥𝐴 ↔ (𝑥𝐴𝑥𝐶)))
32anbi1d 617 . . 3 (𝐴𝐶 → ((𝑥𝐴 ∧ ¬ 𝑥𝐵) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵)))
4 eldif 3779 . . 3 (𝑥 ∈ (𝐴𝐵) ↔ (𝑥𝐴 ∧ ¬ 𝑥𝐵))
5 elin 3995 . . . 4 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ (𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴))
6 eldif 3779 . . . . 5 (𝑥 ∈ (𝐶𝐵) ↔ (𝑥𝐶 ∧ ¬ 𝑥𝐵))
76anbi1i 612 . . . 4 ((𝑥 ∈ (𝐶𝐵) ∧ 𝑥𝐴) ↔ ((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴))
8 ancom 450 . . . . 5 (((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴) ↔ (𝑥𝐴 ∧ (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
9 anass 456 . . . . 5 (((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵) ↔ (𝑥𝐴 ∧ (𝑥𝐶 ∧ ¬ 𝑥𝐵)))
108, 9bitr4i 269 . . . 4 (((𝑥𝐶 ∧ ¬ 𝑥𝐵) ∧ 𝑥𝐴) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵))
115, 7, 103bitri 288 . . 3 (𝑥 ∈ ((𝐶𝐵) ∩ 𝐴) ↔ ((𝑥𝐴𝑥𝐶) ∧ ¬ 𝑥𝐵))
123, 4, 113bitr4g 305 . 2 (𝐴𝐶 → (𝑥 ∈ (𝐴𝐵) ↔ 𝑥 ∈ ((𝐶𝐵) ∩ 𝐴)))
1312eqrdv 2804 1 (𝐴𝐶 → (𝐴𝐵) = ((𝐶𝐵) ∩ 𝐴))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 384   = wceq 1637  wcel 2156  cdif 3766  cin 3768  wss 3769
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2068  ax-7 2104  ax-9 2165  ax-10 2185  ax-11 2201  ax-12 2214  ax-13 2420  ax-ext 2784
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2061  df-clab 2793  df-cleq 2799  df-clel 2802  df-nfc 2937  df-v 3393  df-dif 3772  df-in 3776  df-ss 3783
This theorem is referenced by:  gsumdifsnd  18557  issubdrg  19005  restcld  21187  limcnlp  23855  difelsiga  30520  sigapildsyslem  30548  ldgenpisyslem1  30550  difelcarsg2  30699  ballotlemfp1  30877  asindmre  33805  caragendifcl  41207  gsumdifsndf  42709
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