Theorem u1lemonb 635

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

Assertion

Ref	Expression
u1lemonb	((a →1 b) ∪ b⊥ ) = 1

Proof of Theorem u1lemonb

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 38	. 2 ((a →1 b) ∪ b⊥ ) = ((a⊥ ∪ (a ∩ b)) ∪ b⊥ )
3		or32 82	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ b⊥ ) = ((a⊥ ∪ b⊥ ) ∪ (a ∩ b))
4		df-a 40	. . . . 5 (a ∩ b) = (a⊥ ∪ b⊥ )⊥
5	4	lor 70	. . . 4 ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) = ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ )
6		df-t 41	. . . . 5 1 = ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ )
7	6	ax-r1 35	. . . 4 ((a⊥ ∪ b⊥ ) ∪ (a⊥ ∪ b⊥ )⊥ ) = 1
8	5, 7	ax-r2 36	. . 3 ((a⊥ ∪ b⊥ ) ∪ (a ∩ b)) = 1
9	3, 8	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∪ b⊥ ) = 1
10	2, 9	ax-r2 36	1 ((a →1 b) ∪ b⊥ ) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12

This proof depends on axioms:  ax-a2 31  ax-a3 32  ax-r1 35  ax-r2 36  ax-r5 38

This proof depends on definitions:  df-a 40  df-t 41  df-i1 44

This theorem is used by:  u1lemnab  650  u3lem14a  791