Theorem ud3lem3b 555

Description: Lemma for unified disjunction.

Assertion

Ref	Expression
ud3lem3b	((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = 0

Proof of Theorem ud3lem3b

Step	Hyp	Ref	Expression
1		ud3lem0c 271	. . 3 (a →3 b)⊥ = (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ )))
2	1	ran 71	. 2 ((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)⊥ )
3		an32 76	. . 3 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)⊥ ) = ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ )))
4		anass 69	. . . . . 6 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) = ((a ∪ b⊥ ) ∩ ((a ∪ b) ∩ (a ∪ b)⊥ ))
5		dff 93	. . . . . . . . 9 0 = ((a ∪ b) ∩ (a ∪ b)⊥ )
6	5	ax-r1 34	. . . . . . . 8 ((a ∪ b) ∩ (a ∪ b)⊥ ) = 0
7	6	lan 70	. . . . . . 7 ((a ∪ b⊥ ) ∩ ((a ∪ b) ∩ (a ∪ b)⊥ )) = ((a ∪ b⊥ ) ∩ 0)
8		an0 100	. . . . . . 7 ((a ∪ b⊥ ) ∩ 0) = 0
9	7, 8	ax-r2 35	. . . . . 6 ((a ∪ b⊥ ) ∩ ((a ∪ b) ∩ (a ∪ b)⊥ )) = 0
10	4, 9	ax-r2 35	. . . . 5 (((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) = 0
11	10	ran 71	. . . 4 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = (0 ∩ (a⊥ ∪ (a ∩ b⊥ )))
12		an0r 101	. . . 4 (0 ∩ (a⊥ ∪ (a ∩ b⊥ ))) = 0
13	11, 12	ax-r2 35	. . 3 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ) ∩ (a⊥ ∪ (a ∩ b⊥ ))) = 0
14	3, 13	ax-r2 35	. 2 ((((a ∪ b⊥ ) ∩ (a ∪ b)) ∩ (a⊥ ∪ (a ∩ b⊥ ))) ∩ (a ∪ b)⊥ ) = 0
15	2, 14	ax-r2 35	1 ((a →3 b)⊥ ∩ (a ∪ b)⊥ ) = 0

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  0wf 10   →₃ wi3 15

This theorem is referenced by:  ud3lem3 558

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  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-i3 45

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