Theorem u2lemnanb 656

Description: Lemma for Dishkant implication study. (Contributed by NM, 16-Dec-1997.)

Assertion

Ref	Expression
u2lemnanb	((a →2 b)⊥ ∩ b⊥ ) = ((a ∪ b) ∩ b⊥ )

Proof of Theorem u2lemnanb

Step	Hyp	Ref	Expression
1		u2lemob 631	. . . 4 ((a →2 b) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)
2		anor3 90	. . . . 5 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
3	2	ax-r5 38	. . . 4 ((a⊥ ∩ b⊥ ) ∪ b) = ((a ∪ b)⊥ ∪ b)
4	1, 3	ax-r2 36	. . 3 ((a →2 b) ∪ b) = ((a ∪ b)⊥ ∪ b)
5		oran 87	. . 3 ((a →2 b) ∪ b) = ((a →2 b)⊥ ∩ b⊥ )⊥
6		oran2 92	. . 3 ((a ∪ b)⊥ ∪ b) = ((a ∪ b) ∩ b⊥ )⊥
7	4, 5, 6	3tr2 64	. 2 ((a →2 b)⊥ ∩ b⊥ )⊥ = ((a ∪ b) ∩ b⊥ )⊥
8	7	con1 66	1 ((a →2 b)⊥ ∩ b⊥ ) = ((a ∪ b) ∩ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₂ wi2 13

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-i2 45

This theorem is referenced by: (None)