Theorem u2lemnanb 638

Description: Lemma for Dishkant implication study.

Assertion

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

Proof of Theorem u2lemnanb

Step	Hyp	Ref	Expression
1		u2lemob 613	. . . 4 ((a →2 b) ∪ b) = ((a⊥ ∩ b⊥ ) ∪ b)
2		anor3 82	. . . . 5 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
3	2	ax-r5 37	. . . 4 ((a⊥ ∩ b⊥ ) ∪ b) = ((a ∪ b)⊥ ∪ b)
4	1, 3	ax-r2 35	. . 3 ((a →2 b) ∪ b) = ((a ∪ b)⊥ ∪ b)
5		oran 79	. . 3 ((a →2 b) ∪ b) = ((a →2 b)⊥ ∩ b⊥ )⊥
6		oran2 84	. . 3 ((a ∪ b)⊥ ∪ b) = ((a ∪ b) ∩ b⊥ )⊥
7	4, 5, 6	3tr2 61	. 2 ((a →2 b)⊥ ∩ b⊥ )⊥ = ((a ∪ b) ∩ b⊥ )⊥
8	7	con1 63	1 ((a →2 b)⊥ ∩ b⊥ ) = ((a ∪ b) ∩ b⊥ )

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

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  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-i2 44

metamath.org