Theorem u1lemn1b 712

Description: This theorem continues the line of proofs such as u1lemnaa 622, ud1lem0b 248, u1lemnanb 637, etc. (Contributed by Josiah Burroughs 26-May-04.)

Assertion

Ref	Expression
u1lemn1b	(a →1 b) = ((a →1 b)⊥ →1 b)

Proof of Theorem u1lemn1b

Step	Hyp	Ref	Expression
1		ax-a1 29	. . 3 (a →1 b) = (a →1 b)⊥ ⊥
2		u1lemnab 632	. . . 4 ((a →1 b)⊥ ∩ b) = 0
3	2	ax-r1 34	. . 3 0 = ((a →1 b)⊥ ∩ b)
4	1, 3	2or 67	. 2 ((a →1 b) ∪ 0) = ((a →1 b)⊥ ⊥ ∪ ((a →1 b)⊥ ∩ b))
5		or0 94	. . 3 ((a →1 b) ∪ 0) = (a →1 b)
6	5	ax-r1 34	. 2 (a →1 b) = ((a →1 b) ∪ 0)
7		df-i1 43	. 2 ((a →1 b)⊥ →1 b) = ((a →1 b)⊥ ⊥ ∪ ((a →1 b)⊥ ∩ b))
8	4, 6, 7	3tr1 60	1 (a →1 b) = ((a →1 b)⊥ →1 b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 10   →₁ wi1 13

This theorem is referenced by:  u1lem3var1 713

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-a 39  df-t 40  df-f 41  df-i1 43

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