Theorem u4lemonb 620

Description: Lemma for non-tollens implication study.

Assertion

Ref	Expression
u4lemonb	((a →4 b) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )

Proof of Theorem u4lemonb

Step	Hyp	Ref	Expression
1		df-i4 46	. . 3 (a →4 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ ))
2	1	ax-r5 37	. 2 ((a →4 b) ∪ b⊥ ) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ )
3		ax-a3 31	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ ))
4		lear 153	. . . . 5 ((a⊥ ∪ b) ∩ b⊥ ) ≤ b⊥
5	4	df-le2 123	. . . 4 (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ ) = b⊥
6	5	lor 66	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ )) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )
7	3, 6	ax-r2 35	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )
8	2, 7	ax-r2 35	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 is referenced by:  u4lemnab 635  u4lem3 734

This theorem was proved from axioms:  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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