Theorem u5lemoa 624

Description: Lemma for relevance implication study. (Contributed by NM, 15-Dec-1997.)

Assertion

Ref	Expression
u5lemoa	((a →5 b) ∪ a) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))

Proof of Theorem u5lemoa

Step	Hyp	Ref	Expression
1		df-i5 48	. . 3 (a →5 b) = (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))
2	1	ax-r5 38	. 2 ((a →5 b) ∪ a) = ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ a)
3		ax-a2 31	. . 3 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ a) = (a ∪ (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )))
4		ax-a3 32	. . . . 5 (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) = ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
5	4	lor 70	. . . 4 (a ∪ (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))) = (a ∪ ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))))
6		ax-a3 32	. . . . . 6 ((a ∪ (a ∩ b)) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (a ∪ ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))))
7	6	ax-r1 35	. . . . 5 (a ∪ ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))) = ((a ∪ (a ∩ b)) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
8		orabs 120	. . . . . 6 (a ∪ (a ∩ b)) = a
9	8	ax-r5 38	. . . . 5 ((a ∪ (a ∩ b)) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
10	7, 9	ax-r2 36	. . . 4 (a ∪ ((a ∩ b) ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
11	5, 10	ax-r2 36	. . 3 (a ∪ (((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ ))) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
12	3, 11	ax-r2 36	. 2 ((((a ∩ b) ∪ (a⊥ ∩ b)) ∪ (a⊥ ∩ b⊥ )) ∪ a) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
13	2, 12	ax-r2 36	1 ((a →5 b) ∪ a) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₅ wi5 16

This theorem was proved from axioms:  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r5 38

This theorem depends on definitions:  df-a 40  df-i5 48

This theorem is referenced by:  u5lemnana  649