Theorem u1lemoa 602

Description: Lemma for Sasaki implication study.

Assertion

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

Proof of Theorem u1lemoa

Step	Hyp	Ref	Expression
1		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 37	. 2 ((a →1 b) ∪ a) = ((a⊥ ∪ (a ∩ b)) ∪ a)
3		ax-a2 30	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ a) = (a ∪ (a⊥ ∪ (a ∩ b)))
4		ax-a3 31	. . . . 5 ((a ∪ a⊥ ) ∪ (a ∩ b)) = (a ∪ (a⊥ ∪ (a ∩ b)))
5	4	ax-r1 34	. . . 4 (a ∪ (a⊥ ∪ (a ∩ b))) = ((a ∪ a⊥ ) ∪ (a ∩ b))
6		ax-a2 30	. . . . 5 ((a ∪ a⊥ ) ∪ (a ∩ b)) = ((a ∩ b) ∪ (a ∪ a⊥ ))
7		df-t 40	. . . . . . . 8 1 = (a ∪ a⊥ )
8	7	lor 66	. . . . . . 7 ((a ∩ b) ∪ 1) = ((a ∩ b) ∪ (a ∪ a⊥ ))
9	8	ax-r1 34	. . . . . 6 ((a ∩ b) ∪ (a ∪ a⊥ )) = ((a ∩ b) ∪ 1)
10		or1 96	. . . . . 6 ((a ∩ b) ∪ 1) = 1
11	9, 10	ax-r2 35	. . . . 5 ((a ∩ b) ∪ (a ∪ a⊥ )) = 1
12	6, 11	ax-r2 35	. . . 4 ((a ∪ a⊥ ) ∪ (a ∩ b)) = 1
13	5, 12	ax-r2 35	. . 3 (a ∪ (a⊥ ∪ (a ∩ b))) = 1
14	3, 13	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∪ a) = 1
15	2, 14	ax-r2 35	1 ((a →1 b) ∪ a) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₁ wi1 13

This theorem is referenced by:  u1lemnana 627  oi3oa3lem1 714  i1orni1 829  oau 909

This theorem was proved from axioms:  ax-a2 30  ax-a3 31  ax-a4 32  ax-r1 34  ax-r2 35  ax-r5 37

This theorem depends on definitions:  df-t 40  df-i1 43

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