Theorem di 126

Description: Dishkant implication.

Assertion

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

Proof of Theorem di

Step	Hyp	Ref	Expression
1		conb 122	. . 3 ((b⊥ ∪ a⊥ )⊥ ≡ a) = ((b⊥ ∪ a⊥ )⊥ ⊥ ≡ a⊥ )
2		ax-a1 30	. . . . . 6 (b⊥ ∪ a⊥ ) = (b⊥ ∪ a⊥ )⊥ ⊥
3	2	ax-r1 35	. . . . 5 (b⊥ ∪ a⊥ )⊥ ⊥ = (b⊥ ∪ a⊥ )
4	3	rbi 98	. . . 4 ((b⊥ ∪ a⊥ )⊥ ⊥ ≡ a⊥ ) = ((b⊥ ∪ a⊥ ) ≡ a⊥ )
5		mi 125	. . . 4 ((b⊥ ∪ a⊥ ) ≡ a⊥ ) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
6	4, 5	ax-r2 36	. . 3 ((b⊥ ∪ a⊥ )⊥ ⊥ ≡ a⊥ ) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
7	1, 6	ax-r2 36	. 2 ((b⊥ ∪ a⊥ )⊥ ≡ a) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
8		ancom 74	. . . 4 (a ∩ b) = (b ∩ a)
9		df-a 40	. . . 4 (b ∩ a) = (b⊥ ∪ a⊥ )⊥
10	8, 9	ax-r2 36	. . 3 (a ∩ b) = (b⊥ ∪ a⊥ )⊥
11	10	rbi 98	. 2 ((a ∩ b) ≡ a) = ((b⊥ ∪ a⊥ )⊥ ≡ a)
12		ax-a1 30	. . . . 5 b = b⊥ ⊥
13		ax-a1 30	. . . . 5 a = a⊥ ⊥
14	12, 13	2an 79	. . . 4 (b ∩ a) = (b⊥ ⊥ ∩ a⊥ ⊥ )
15	8, 14	ax-r2 36	. . 3 (a ∩ b) = (b⊥ ⊥ ∩ a⊥ ⊥ )
16	15	lor 70	. 2 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (b⊥ ⊥ ∩ a⊥ ⊥ ))
17	7, 11, 16	3tr1 63	1 ((a ∩ b) ≡ a) = (a⊥ ∪ (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: (None)