Theorem u2lemnana 628

Description: Lemma for Dishkant implication study.

Assertion

Ref	Expression
u2lemnana	((a →2 b)⊥ ∩ a⊥ ) = 0

Proof of Theorem u2lemnana

Step	Hyp	Ref	Expression
1		anor3 82	. . 3 ((a →2 b)⊥ ∩ a⊥ ) = ((a →2 b) ∪ a)⊥
2		u2lemoa 603	. . . 4 ((a →2 b) ∪ a) = 1
3	2	ax-r4 36	. . 3 ((a →2 b) ∪ a)⊥ = 1⊥
4	1, 3	ax-r2 35	. 2 ((a →2 b)⊥ ∩ a⊥ ) = 1⊥
5		df-f 41	. . 3 0 = 1⊥
6	5	ax-r1 34	. 2 1⊥ = 0
7	4, 6	ax-r2 35	1 ((a →2 b)⊥ ∩ a⊥ ) = 0

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9  0wf 10   →₂ wi2 14

This theorem is referenced by:  u2lem5 744

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  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

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