Theorem u1lemnaa 640

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

Assertion

Ref	Expression
u1lemnaa	((a →1 b)⊥ ∩ a) = (a ∩ (a⊥ ∪ b⊥ ))

Proof of Theorem u1lemnaa

Step	Hyp	Ref	Expression
1		anor2 89	. 2 ((a →1 b)⊥ ∩ a) = ((a →1 b) ∪ a⊥ )⊥
2		u1lemona 625	. . . 4 ((a →1 b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
3	2	ax-r4 37	. . 3 ((a →1 b) ∪ a⊥ )⊥ = (a⊥ ∪ (a ∩ b))⊥
4		df-a 40	. . . . 5 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥
5		df-a 40	. . . . . . . 8 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
6	5	lor 70	. . . . . . 7 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
7	6	ax-r4 37	. . . . . 6 (a⊥ ∪ (a ∩ b))⊥ = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥
8	7	ax-r1 35	. . . . 5 (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥ = (a⊥ ∪ (a ∩ b))⊥
9	4, 8	ax-r2 36	. . . 4 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a ∩ b))⊥
10	9	ax-r1 35	. . 3 (a⊥ ∪ (a ∩ b))⊥ = (a ∩ (a⊥ ∪ b⊥ ))
11	3, 10	ax-r2 36	. 2 ((a →1 b) ∪ a⊥ )⊥ = (a ∩ (a⊥ ∪ b⊥ ))
12	1, 11	ax-r2 36	1 ((a →1 b)⊥ ∩ a) = (a ∩ (a⊥ ∪ b⊥ ))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

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-i1 44

This theorem is referenced by: (None)