Theorem ud2lem0c 270

Description: Lemma for unified disjunction.

Assertion

Ref	Expression
ud2lem0c	(a →2 b)⊥ = (b⊥ ∩ (a ∪ b))

Proof of Theorem ud2lem0c

Step	Hyp	Ref	Expression
1		df-i2 44	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2		oran 79	. . . 4 (b ∪ (a⊥ ∩ b⊥ )) = (b⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥
3		oran 79	. . . . . . 7 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
4	3	ax-r1 34	. . . . . 6 (a⊥ ∩ b⊥ )⊥ = (a ∪ b)
5	4	lan 70	. . . . 5 (b⊥ ∩ (a⊥ ∩ b⊥ )⊥ ) = (b⊥ ∩ (a ∪ b))
6	5	ax-r4 36	. . . 4 (b⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥ = (b⊥ ∩ (a ∪ b))⊥
7	2, 6	ax-r2 35	. . 3 (b ∪ (a⊥ ∩ b⊥ )) = (b⊥ ∩ (a ∪ b))⊥
8	1, 7	ax-r2 35	. 2 (a →2 b) = (b⊥ ∩ (a ∪ b))⊥
9	8	con2 64	1 (a →2 b)⊥ = (b⊥ ∩ (a ∪ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 14

This theorem is referenced by:  wql2lem5 284  ud2lem1 545  ud2lem3 547  u2lem1 717  3vth9 794  2oalem1 807  oa43v 1008  oa63v 1011

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-i2 44

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