Theorem u5lemnab 654

Description: Lemma for relevance implication study. (Contributed by NM, 16-Dec-1997.)

Assertion

Ref	Expression
u5lemnab	((a →5 b)⊥ ∩ b) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)

Proof of Theorem u5lemnab

Step	Hyp	Ref	Expression
1		u5lemonb 639	. . . 4 ((a →5 b) ∪ b⊥ ) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ )
2		ax-a2 31	. . . . . 6 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a⊥ ∩ b) ∪ (a ∩ b))
3		anor2 89	. . . . . . . 8 (a⊥ ∩ b) = (a ∪ b⊥ )⊥
4		df-a 40	. . . . . . . 8 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	3, 4	2or 72	. . . . . . 7 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ )
6		oran3 93	. . . . . . 7 ((a ∪ b⊥ )⊥ ∪ (a⊥ ∪ b⊥ )⊥ ) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
7	5, 6	ax-r2 36	. . . . . 6 ((a⊥ ∩ b) ∪ (a ∩ b)) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
8	2, 7	ax-r2 36	. . . . 5 ((a ∩ b) ∪ (a⊥ ∩ b)) = ((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥
9	8	ax-r5 38	. . . 4 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ b⊥ ) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥ ∪ b⊥ )
10	1, 9	ax-r2 36	. . 3 ((a →5 b) ∪ b⊥ ) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥ ∪ b⊥ )
11		oran1 91	. . 3 ((a →5 b) ∪ b⊥ ) = ((a →5 b)⊥ ∩ b)⊥
12		oran3 93	. . 3 (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ ))⊥ ∪ b⊥ ) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)⊥
13	10, 11, 12	3tr2 64	. 2 ((a →5 b)⊥ ∩ b)⊥ = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)⊥
14	13	con1 66	1 ((a →5 b)⊥ ∩ b) = (((a ∪ b⊥ ) ∩ (a⊥ ∪ b⊥ )) ∩ b)

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 16

This proof depends on 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 proof depends on definitions:  df-a 40  df-i5 48  df-le1 130  df-le2 131

This theorem is used by: (None)