Theorem u4lemnab 635

Description: Lemma for non-tollens implication study.

Assertion

Ref	Expression
u4lemnab	((a →4 b)⊥ ∩ b) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)

Proof of Theorem u4lemnab

Step	Hyp	Ref	Expression
1		u4lemonb 620	. . . 4 ((a →4 b) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )
2		ax-a2 30	. . . . . 6 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a ∩ b))
3		anor2 81	. . . . . . . 8 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
4		df-a 39	. . . . . . . 8 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	3, 4	2or 67	. . . . . . 7 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
6		oran3 85	. . . . . . 7 ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ ) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
7	5, 6	ax-r2 35	. . . . . 6 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
8	2, 7	ax-r2 35	. . . . 5 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
9	8	ax-r5 37	. . . 4 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ ) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥ ∪ b⊥ )
10	1, 9	ax-r2 35	. . 3 ((a →4 b) ∪ b⊥ ) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥ ∪ b⊥ )
11		oran1 83	. . 3 ((a →4 b) ∪ b⊥ ) = ((a →4 b)⊥ ∩ b)⊥
12		oran3 85	. . 3 (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥ ∪ b⊥ ) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)⊥
13	10, 11, 12	3tr2 61	. 2 ((a →4 b)⊥ ∩ b)⊥ = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)⊥
14	13	con1 63	1 ((a →4 b)⊥ ∩ b) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 16

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-i4 46  df-le1 122  df-le2 123

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