Theorem ud2lem0c 278

Description: Lemma for unified disjunction. (Contributed by NM, 23-Nov-1997.)

Assertion

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

Proof of Theorem ud2lem0c

Step	Hyp	Ref	Expression
1		df-i2 45	. . 3 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2		oran 87	. . . 4 (b ∪ (a⊥ ∩ b⊥ )) = (b⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥
3		oran 87	. . . . . . 7 (a ∪ b) = (a⊥ ∩ b⊥ )⊥
4	3	ax-r1 35	. . . . . 6 (a⊥ ∩ b⊥ )⊥ = (a ∪ b)
5	4	lan 77	. . . . 5 (b⊥ ∩ (a⊥ ∩ b⊥ )⊥ ) = (b⊥ ∩ (a ∪ b))
6	5	ax-r4 37	. . . 4 (b⊥ ∩ (a⊥ ∩ b⊥ )⊥ )⊥ = (b⊥ ∩ (a ∪ b))⊥
7	2, 6	ax-r2 36	. . 3 (b ∪ (a⊥ ∩ b⊥ )) = (b⊥ ∩ (a ∪ b))⊥
8	1, 7	ax-r2 36	. 2 (a →2 b) = (b⊥ ∩ (a ∪ b))⊥
9	8	con2 67	1 (a →2 b)⊥ = (b⊥ ∩ (a ∪ b))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 13

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i2 45

This theorem is used by:  wql2lem5  292  ud2lem1  563  ud2lem3  565  u2lem1  735  3vth9  812  2oalem1  825  oa43v  1028  oa63v  1032