Theorem u1lemnaa 622

Description: Lemma for Sasaki implication study.

Assertion

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

Proof of Theorem u1lemnaa

Step	Hyp	Ref	Expression
1		anor2 81	. 2 ((a →1 b)⊥ ∩ a) = ((a →1 b) ∪ a⊥ )⊥
2		u1lemona 607	. . . 4 ((a →1 b) ∪ a⊥ ) = (a⊥ ∪ (a ∩ b))
3	2	ax-r4 36	. . 3 ((a →1 b) ∪ a⊥ )⊥ = (a⊥ ∪ (a ∩ b))⊥
4		df-a 39	. . . . 5 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥
5		df-a 39	. . . . . . . 8 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
6	5	lor 66	. . . . . . 7 (a⊥ ∪ (a ∩ b)) = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
7	6	ax-r4 36	. . . . . 6 (a⊥ ∪ (a ∩ b))⊥ = (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥
8	7	ax-r1 34	. . . . 5 (a⊥ ∪ (a⊥ ∪ b⊥ )⊥ )⊥ = (a⊥ ∪ (a ∩ b))⊥
9	4, 8	ax-r2 35	. . . 4 (a ∩ (a⊥ ∪ b⊥ )) = (a⊥ ∪ (a ∩ b))⊥
10	9	ax-r1 34	. . 3 (a⊥ ∪ (a ∩ b))⊥ = (a ∩ (a⊥ ∪ b⊥ ))
11	3, 10	ax-r2 35	. 2 ((a →1 b) ∪ a⊥ )⊥ = (a ∩ (a⊥ ∪ b⊥ ))
12	1, 11	ax-r2 35	1 ((a →1 b)⊥ ∩ a) = (a ∩ (a⊥ ∪ b⊥ ))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

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-i1 43

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