Theorem u1lemoa 620

Description: Lemma for Sasaki implication study. (Contributed by NM, 14-Dec-1997.)

Assertion

Ref	Expression
u1lemoa	((a →1 b) ∪ a) = 1

Proof of Theorem u1lemoa

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 38	. 2 ((a →1 b) ∪ a) = ((a⊥ ∪ (a ∩ b)) ∪ a)
3		ax-a2 31	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ a) = (a ∪ (a⊥ ∪ (a ∩ b)))
4		ax-a3 32	. . . . 5 ((a ∪ a⊥ ) ∪ (a ∩ b)) = (a ∪ (a⊥ ∪ (a ∩ b)))
5	4	ax-r1 35	. . . 4 (a ∪ (a⊥ ∪ (a ∩ b))) = ((a ∪ a⊥ ) ∪ (a ∩ b))
6		ax-a2 31	. . . . 5 ((a ∪ a⊥ ) ∪ (a ∩ b)) = ((a ∩ b) ∪ (a ∪ a⊥ ))
7		df-t 41	. . . . . . . 8 1 = (a ∪ a⊥ )
8	7	lor 70	. . . . . . 7 ((a ∩ b) ∪ 1) = ((a ∩ b) ∪ (a ∪ a⊥ ))
9	8	ax-r1 35	. . . . . 6 ((a ∩ b) ∪ (a ∪ a⊥ )) = ((a ∩ b) ∪ 1)
10		or1 104	. . . . . 6 ((a ∩ b) ∪ 1) = 1
11	9, 10	ax-r2 36	. . . . 5 ((a ∩ b) ∪ (a ∪ a⊥ )) = 1
12	6, 11	ax-r2 36	. . . 4 ((a ∪ a⊥ ) ∪ (a ∩ b)) = 1
13	5, 12	ax-r2 36	. . 3 (a ∪ (a⊥ ∪ (a ∩ b))) = 1
14	3, 13	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∪ a) = 1
15	2, 14	ax-r2 36	1 ((a →1 b) ∪ a) = 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-a4 33  ax-r1 35  ax-r2 36  ax-r5 38

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

This theorem is used by:  u1lemnana  645  oi3oa3lem1  732  i1orni1  847  oau  929