Theorem biao 781

Description: Equivalence to biconditional.

Assertion

Ref	Expression
biao	(a ≡ b) = ((a ∩ b) ≡ (a ∪ b))

Proof of Theorem biao

Step	Hyp	Ref	Expression
1		leao1 154	. . . . 5 (a ∩ b) ≤ (a ∪ b)
2	1	df2le2 128	. . . 4 ((a ∩ b) ∩ (a ∪ b)) = (a ∩ b)
3	2	ax-r1 34	. . 3 (a ∩ b) = ((a ∩ b) ∩ (a ∪ b))
4		anor3 82	. . . 4 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
5	1	lecon 146	. . . . . 6 (a ∪ b)⊥ ≤ (a ∩ b)⊥
6		oridm 102	. . . . . . 7 ((a ∪ b)⊥ ∪ (a ∪ b)⊥ ) = (a ∪ b)⊥
7	6	df-le1 122	. . . . . 6 (a ∪ b)⊥ ≤ (a ∪ b)⊥
8	5, 7	ler2an 165	. . . . 5 (a ∪ b)⊥ ≤ ((a ∩ b)⊥ ∩ (a ∪ b)⊥ )
9		lear 153	. . . . . . 7 ((a ∩ b)⊥ ∩ (a ∪ b)⊥ ) ≤ (a ∪ b)⊥
10	9	df-le2 123	. . . . . 6 (((a ∩ b)⊥ ∩ (a ∪ b)⊥ ) ∪ (a ∪ b)⊥ ) = (a ∪ b)⊥
11	10	df-le1 122	. . . . 5 ((a ∩ b)⊥ ∩ (a ∪ b)⊥ ) ≤ (a ∪ b)⊥
12	8, 11	lebi 137	. . . 4 (a ∪ b)⊥ = ((a ∩ b)⊥ ∩ (a ∪ b)⊥ )
13	4, 12	ax-r2 35	. . 3 (a⊥ ∩ b⊥ ) = ((a ∩ b)⊥ ∩ (a ∪ b)⊥ )
14	3, 13	2or 67	. 2 ((a ∩ b) ∪ (a⊥ ∩ b⊥ )) = (((a ∩ b) ∩ (a ∪ b)) ∪ ((a ∩ b)⊥ ∩ (a ∪ b)⊥ ))
15		dfb 86	. 2 (a ≡ b) = ((a ∩ b) ∪ (a⊥ ∩ b⊥ ))
16		dfb 86	. 2 ((a ∩ b) ≡ (a ∪ b)) = (((a ∩ b) ∩ (a ∪ b)) ∪ ((a ∩ b)⊥ ∩ (a ∪ b)⊥ ))
17	14, 15, 16	3tr1 60	1 (a ≡ b) = ((a ∩ b) ≡ (a ∪ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ≡ tb 5   ∪ wo 6   ∩ wa 7

This theorem is referenced by:  mlaconj4 826

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-le1 122  df-le2 123

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