Theorem lem3.3.7i1e1 1060

Description: Equation 3.7 of [PavMeg1999] p. 9. The variable i in the paper is set to 1, and this is the first part of the equation. (Contributed by Roy F. Longton, 3-Jul-2005.)

Assertion

Ref	Expression
lem3.3.7i1e1	(a →1 (a ∩ b)) = (a ≡1 (a ∩ b))

Proof of Theorem lem3.3.7i1e1

Step	Hyp	Ref	Expression
1		or1r 105	. . . . . 6 (1 ∪ b⊥ ) = 1
2	1	ax-r1 35	. . . . 5 1 = (1 ∪ b⊥ )
3	2	ran 78	. . . 4 (1 ∩ (a⊥ ∪ (a ∩ (a ∩ b)))) = ((1 ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
4		an1r 107	. . . 4 (1 ∩ (a⊥ ∪ (a ∩ (a ∩ b)))) = (a⊥ ∪ (a ∩ (a ∩ b)))
5		df-t 41	. . . . . 6 1 = (a ∪ a⊥ )
6	5	ax-r5 38	. . . . 5 (1 ∪ b⊥ ) = ((a ∪ a⊥ ) ∪ b⊥ )
7	6	ran 78	. . . 4 ((1 ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b)))) = (((a ∪ a⊥ ) ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
8	3, 4, 7	3tr2 64	. . 3 (a⊥ ∪ (a ∩ (a ∩ b))) = (((a ∪ a⊥ ) ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
9		ax-a3 32	. . . 4 ((a ∪ a⊥ ) ∪ b⊥ ) = (a ∪ (a⊥ ∪ b⊥ ))
10	9	ran 78	. . 3 (((a ∪ a⊥ ) ∪ b⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b)))) = ((a ∪ (a⊥ ∪ b⊥ )) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
11		oran3 93	. . . . 5 (a⊥ ∪ b⊥ ) = (a ∩ b)⊥
12	11	lor 70	. . . 4 (a ∪ (a⊥ ∪ b⊥ )) = (a ∪ (a ∩ b)⊥ )
13	12	ran 78	. . 3 ((a ∪ (a⊥ ∪ b⊥ )) ∩ (a⊥ ∪ (a ∩ (a ∩ b)))) = ((a ∪ (a ∩ b)⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
14	8, 10, 13	3tr 65	. 2 (a⊥ ∪ (a ∩ (a ∩ b))) = ((a ∪ (a ∩ b)⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
15		df-i1 44	. 2 (a →1 (a ∩ b)) = (a⊥ ∪ (a ∩ (a ∩ b)))
16		df-id1 50	. 2 (a ≡1 (a ∩ b)) = ((a ∪ (a ∩ b)⊥ ) ∩ (a⊥ ∪ (a ∩ (a ∩ b))))
17	14, 15, 16	3tr1 63	1 (a →1 (a ∩ b)) = (a ≡1 (a ∩ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12   ≡₁ wid1 18

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44  df-id1 50

This theorem is referenced by: (None)