Theorem u4lemonb 638

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 47	. . 3 (a →4 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ ))
2	1	ax-r5 38	. 2 ((a →4 b) ∪ b⊥ ) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ )
3		ax-a3 32	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ ))
4		lear 161	. . . . 5 ((a⊥ ∪ b) ∩ b⊥ ) ≤ b⊥
5	4	df-le2 131	. . . 4 (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ ) = b⊥
6	5	lor 70	. . 3 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (((a⊥ ∪ b) ∩ b⊥ ) ∪ b⊥ )) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )
7	3, 6	ax-r2 36	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ ((a⊥ ∪ b) ∩ b⊥ )) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )
8	2, 7	ax-r2 36	1 ((a →4 b) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₄ wi4 15

This theorem was proved from axioms:  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-i4 47  df-le1 130  df-le2 131

This theorem is referenced by:  u4lemnab  653  u4lem3  752