Theorem mi 125

Description: Mittelstaedt implication.

Assertion

Ref	Expression
mi	((a ∪ b) ≡ b) = (b ∪ (a⊥ ∩ b⊥ ))

Proof of Theorem mi

Step	Hyp	Ref	Expression
1		dfb 94	. 2 ((a ∪ b) ≡ b) = (((a ∪ b) ∩ b) ∪ ((a ∪ b)⊥ ∩ b⊥ ))
2		ancom 74	. . . 4 ((a ∪ b) ∩ b) = (b ∩ (a ∪ b))
3		ax-a2 31	. . . . . 6 (a ∪ b) = (b ∪ a)
4	3	lan 77	. . . . 5 (b ∩ (a ∪ b)) = (b ∩ (b ∪ a))
5		anabs 121	. . . . 5 (b ∩ (b ∪ a)) = b
6	4, 5	ax-r2 36	. . . 4 (b ∩ (a ∪ b)) = b
7	2, 6	ax-r2 36	. . 3 ((a ∪ b) ∩ b) = b
8		oran 87	. . . . . . 7 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
9	8	con2 67	. . . . . 6 (a ∪ b)⊥ = (a⊥ ∩ b⊥ )
10	9	ran 78	. . . . 5 ((a ∪ b)⊥ ∩ b⊥ ) = ((a⊥ ∩ b⊥ ) ∩ b⊥ )
11		anass 76	. . . . 5 ((a⊥ ∩ b⊥ ) ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ ))
12	10, 11	ax-r2 36	. . . 4 ((a ∪ b)⊥ ∩ b⊥ ) = (a⊥ ∩ (b⊥ ∩ b⊥ ))
13		anidm 111	. . . . 5 (b⊥ ∩ b⊥ ) = b⊥
14	13	lan 77	. . . 4 (a⊥ ∩ (b⊥ ∩ b⊥ )) = (a⊥ ∩ b⊥ )
15	12, 14	ax-r2 36	. . 3 ((a ∪ b)⊥ ∩ b⊥ ) = (a⊥ ∩ b⊥ )
16	7, 15	2or 72	. 2 (((a ∪ b) ∩ b) ∪ ((a ∪ b)⊥ ∩ b⊥ )) = (b ∪ (a⊥ ∩ b⊥ ))
17	1, 16	ax-r2 36	1 ((a ∪ b) ≡ b) = (b ∪ (a⊥ ∩ b⊥ ))

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

This theorem is referenced by:  di  126  lei2  346