Theorem ud4lem3a 583

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 76	. . 3 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∩ (a ∪ b)) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (((a ∩ b⊥ ) ∪ b) ∩ (a ∪ b)))
2		lea 160	. . . . . 6 (a ∩ b⊥ ) ≤ a
3	2	leror 152	. . . . 5 ((a ∩ b⊥ ) ∪ b) ≤ (a ∪ b)
4	3	df2le2 136	. . . 4 (((a ∩ b⊥ ) ∪ b) ∩ (a ∪ b)) = ((a ∩ b⊥ ) ∪ b)
5	4	lan 77	. . 3 (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ (((a ∩ b⊥ ) ∪ b) ∩ (a ∪ b))) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
6	1, 5	ax-r2 36	. 2 ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∩ (a ∪ b)) = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
7		ud4lem0c 280	. . 3 (a →4 b)⊥ = (((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b))
8	7	ran 78	. 2 ((a →4 b)⊥ ∩ (a ∪ b)) = ((((a⊥ ∪ b⊥ ) ∩ (a ∪ b⊥ )) ∩ ((a ∩ b⊥ ) ∪ b)) ∩ (a ∪ b))
9	6, 8, 7	3tr1 63	1 ((a →4 b)⊥ ∩ (a ∪ b)) = (a →4 b)⊥

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 15

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-a 40  df-t 41  df-f 42  df-i4 47  df-le1 130  df-le2 131

This theorem is referenced by:  ud4lem3  585