Theorem lem3.3.5 1055

Description: Equation 3.5 of [PavMeg1999] p. 9. (Contributed by Roy F. Longton, 28-Jun-2005.) (Revised by Roy F. Longton, 3-Jul-2005.)

Hypothesis

Ref	Expression
lem3.3.5.1	(a ≡5 b) = 1

Assertion

Ref	Expression
lem3.3.5	(a →1 (b ∪ c)) = 1

Proof of Theorem lem3.3.5

Step	Hyp	Ref	Expression
1		df-b1 1048	. . . . . 6 (a ↔1 b) = ((a →1 b) ∩ (b →1 a))
2		lea 160	. . . . . 6 ((a →1 b) ∩ (b →1 a)) ≤ (a →1 b)
3	1, 2	bltr 138	. . . . 5 (a ↔1 b) ≤ (a →1 b)
4		df-i1 44	. . . . 5 (a →1 b) = (a⊥ ∪ (a ∩ b))
5	3, 4	lbtr 139	. . . 4 (a ↔1 b) ≤ (a⊥ ∪ (a ∩ b))
6		leo 158	. . . . . 6 b ≤ (b ∪ c)
7	6	lelan 167	. . . . 5 (a ∩ b) ≤ (a ∩ (b ∪ c))
8	7	lelor 166	. . . 4 (a⊥ ∪ (a ∩ b)) ≤ (a⊥ ∪ (a ∩ (b ∪ c)))
9	5, 8	letr 137	. . 3 (a ↔1 b) ≤ (a⊥ ∪ (a ∩ (b ∪ c)))
10		lem3.3.5.1	. . . . 5 (a ≡5 b) = 1
11		lem3.3.3 1052	. . . . 5 ((a ≡5 b) →0 (a ↔1 b)) = 1
12	10, 11	lem3.3.2 1046	. . . 4 (a ↔1 b) = 1
13	12	ax-r1 35	. . 3 1 = (a ↔1 b)
14		df-i1 44	. . 3 (a →1 (b ∪ c)) = (a⊥ ∪ (a ∩ (b ∪ c)))
15	9, 13, 14	le3tr1 140	. 2 1 ≤ (a →1 (b ∪ c))
16	15	lem3.3.5lem 1054	1 (a →1 (b ∪ c)) = 1

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12   ≡₅ wid5 22   ↔₁ wb1 24

This proof depends on axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This proof depends on definitions:  df-a 40  df-t 41  df-f 42  df-i0 43  df-i1 44  df-le1 130  df-le2 131  df-id5 1047  df-b1 1048

This theorem is used by:  lem3.4.5  1078