Theorem ud4lem3a 565

Description: Lemma for unified disjunction.

Assertion

Ref	Expression
ud4lem3a	((a →4 b)⊥ ∩ (a ∪ b)) = (a →4 b)⊥

Proof of Theorem ud4lem3a

Step	Hyp	Ref	Expression
1		anass 69	. . 3 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∩ (a ∪ b)) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (((a ∩ b⊥ ) ∪ b) ∩ (a ∪ b)))
2		lea 152	. . . . . 6 (a ∩ b⊥ ) ≤ a
3	2	leror 144	. . . . 5 ((a ∩ b⊥ ) ∪ b) ≤ (a ∪ b)
4	3	df2le2 128	. . . 4 (((a ∩ b⊥ ) ∪ b) ∩ (a ∪ b)) = ((a ∩ b⊥ ) ∪ b)
5	4	lan 70	. . 3 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (((a ∩ b⊥ ) ∪ b) ∩ (a ∪ b))) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
6	1, 5	ax-r2 35	. 2 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∩ (a ∪ b)) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
7		ud4lem0c 272	. . 3 (a →4 b)⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
8	7	ran 71	. 2 ((a →4 b)⊥ ∩ (a ∪ b)) = ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∩ (a ∪ b))
9	6, 8, 7	3tr1 60	1 ((a →4 b)⊥ ∩ (a ∪ b)) = (a →4 b)⊥

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 16

This theorem is referenced by:  ud4lem3 567

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-i4 46  df-le1 122  df-le2 123

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