Theorem u1lemnana 645

Description: Lemma for Sasaki implication study. (Contributed by NM, 15-Dec-1997.)

Assertion

Ref	Expression
u1lemnana	((a →1 b)⊥ ∩ a⊥ ) = 0

Proof of Theorem u1lemnana

Step	Hyp	Ref	Expression
1		anor3 90	. . 3 ((a →1 b)⊥ ∩ a⊥ ) = ((a →1 b) ∪ a)⊥
2		u1lemoa 620	. . . 4 ((a →1 b) ∪ a) = 1
3	2	ax-r4 37	. . 3 ((a →1 b) ∪ a)⊥ = 1⊥
4	1, 3	ax-r2 36	. 2 ((a →1 b)⊥ ∩ a⊥ ) = 1⊥
5		df-f 42	. . 3 0 = 1⊥
6	5	ax-r1 35	. 2 1⊥ = 0
7	4, 6	ax-r2 36	1 ((a →1 b)⊥ ∩ a⊥ ) = 0

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8  0wf 9   →₁ wi1 12

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  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-i1 44

This theorem is referenced by: (None)