Theorem u1lemob 630

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

Assertion

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

Proof of Theorem u1lemob

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		ax-a2 31	. . . 4 ((a⊥ ∪ b) ∪ (a ∩ b)) = ((a ∩ b) ∪ (a⊥ ∪ b))
5		lear 161	. . . . . 6 (a ∩ b) ≤ b
6		leor 159	. . . . . 6 b ≤ (a⊥ ∪ b)
7	5, 6	letr 137	. . . . 5 (a ∩ b) ≤ (a⊥ ∪ b)
8	7	df-le2 131	. . . 4 ((a ∩ b) ∪ (a⊥ ∪ b)) = (a⊥ ∪ b)
9	4, 8	ax-r2 36	. . 3 ((a⊥ ∪ b) ∪ (a ∩ b)) = (a⊥ ∪ b)
10	3, 9	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∪ b) = (a⊥ ∪ b)
11	2, 10	ax-r2 36	1 ((a →1 b) ∪ b) = (a⊥ ∪ b)

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

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-i1 44  df-le1 130  df-le2 131

This theorem is used by:  u1lemnanb  655  u12lem  771  salem1  837