Theorem u1lemn1b 730

Description: This theorem continues the line of proofs such as u1lemnaa 640, ud1lem0b 256, u1lemnanb 655, etc. (Contributed by Josiah Burroughs, 26-May-2004.)

Assertion

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

Proof of Theorem u1lemn1b

Step	Hyp	Ref	Expression
1		ax-a1 30	. . 3 (a →1 b) = (a →1 b)⊥ ⊥
2		u1lemnab 650	. . . 4 ((a →1 b)⊥ ∩ b) = 0
3	2	ax-r1 35	. . 3 0 = ((a →1 b)⊥ ∩ b)
4	1, 3	2or 72	. 2 ((a →1 b) ∪ 0) = ((a →1 b)⊥ ⊥ ∪ ((a →1 b)⊥ ∩ b))
5		or0 102	. . 3 ((a →1 b) ∪ 0) = (a →1 b)
6	5	ax-r1 35	. 2 (a →1 b) = ((a →1 b) ∪ 0)
7		df-i1 44	. 2 ((a →1 b)⊥ →1 b) = ((a →1 b)⊥ ⊥ ∪ ((a →1 b)⊥ ∩ b))
8	4, 6, 7	3tr1 63	1 (a →1 b) = ((a →1 b)⊥ →1 b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 9   →₁ wi1 12

This proof depends on 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 proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-i1 44

This theorem is used by:  u1lem3var1  731  lem4.6.5  1087