Theorem u1lemob 612

Description: Lemma for Sasaki implication study.

Assertion

Ref	Expression
u1lemob	((a →1 b) ∪ b) = (a⊥ ∪ b)

Proof of Theorem u1lemob

Step	Hyp	Ref	Expression
1		df-i1 43	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ax-r5 37	. 2 ((a →1 b) ∪ b) = ((a⊥ ∪ (a ∩ b)) ∪ b)
3		or32 75	. . 3 ((a⊥ ∪ (a ∩ b)) ∪ b) = ((a⊥ ∪ b) ∪ (a ∩ b))
4		ax-a2 30	. . . 4 ((a⊥ ∪ b) ∪ (a ∩ b)) = ((a ∩ b) ∪ (a⊥ ∪ b))
5		lear 153	. . . . . 6 (a ∩ b) ≤ b
6		leor 151	. . . . . 6 b ≤ (a⊥ ∪ b)
7	5, 6	letr 129	. . . . 5 (a ∩ b) ≤ (a⊥ ∪ b)
8	7	df-le2 123	. . . 4 ((a ∩ b) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b)
9	4, 8	ax-r2 35	. . 3 ((a⊥ ∪ b) ∪ (a ∩ b)) = (a⊥ ∪ b)
10	3, 9	ax-r2 35	. 2 ((a⊥ ∪ (a ∩ b)) ∪ b) = (a⊥ ∪ b)
11	2, 10	ax-r2 35	1 ((a →1 b) ∪ b) = (a⊥ ∪ b)

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

This theorem is referenced by:  u1lemnanb 637  u12lem 753  salem1 819

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-i1 43  df-le1 122  df-le2 123

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