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Theorem biao 799
 Description: Equivalence to biconditional. (Contributed by NM, 8-Jul-2000.)
Assertion
Ref Expression
biao (ab) = ((ab) ≡ (ab))

Proof of Theorem biao
StepHypRef Expression
1 leao1 162 . . . . 5 (ab) ≤ (ab)
21df2le2 136 . . . 4 ((ab) ∩ (ab)) = (ab)
32ax-r1 35 . . 3 (ab) = ((ab) ∩ (ab))
4 anor3 90 . . . 4 (ab ) = (ab)
51lecon 154 . . . . . 6 (ab) ≤ (ab)
6 oridm 110 . . . . . . 7 ((ab) ∪ (ab) ) = (ab)
76df-le1 130 . . . . . 6 (ab) ≤ (ab)
85, 7ler2an 173 . . . . 5 (ab) ≤ ((ab) ∩ (ab) )
9 lear 161 . . . . . . 7 ((ab) ∩ (ab) ) ≤ (ab)
109df-le2 131 . . . . . 6 (((ab) ∩ (ab) ) ∪ (ab) ) = (ab)
1110df-le1 130 . . . . 5 ((ab) ∩ (ab) ) ≤ (ab)
128, 11lebi 145 . . . 4 (ab) = ((ab) ∩ (ab) )
134, 12ax-r2 36 . . 3 (ab ) = ((ab) ∩ (ab) )
143, 132or 72 . 2 ((ab) ∪ (ab )) = (((ab) ∩ (ab)) ∪ ((ab) ∩ (ab) ))
15 dfb 94 . 2 (ab) = ((ab) ∪ (ab ))
16 dfb 94 . 2 ((ab) ≡ (ab)) = (((ab) ∩ (ab)) ∪ ((ab) ∩ (ab) ))
1714, 15, 163tr1 63 1 (ab) = ((ab) ≡ (ab))
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-le1 130  df-le2 131 This theorem is referenced by:  mlaconj4  844
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