Theorem i1or 345

Description: Lemma for disjunction of →₁ . (Contributed by NM, 5-Jul-2000.)

Assertion

Ref	Expression
i1or	((c →1 a) ∪ (c →1 b)) ≤ (c →1 (a ∪ b))

Proof of Theorem i1or

Step	Hyp	Ref	Expression
1		df-i1 44	. . . 4 (c →1 a) = (c⊥ ∪ (c ∩ a))
2		leo 158	. . . . . 6 a ≤ (a ∪ b)
3	2	lelan 167	. . . . 5 (c ∩ a) ≤ (c ∩ (a ∪ b))
4	3	lelor 166	. . . 4 (c⊥ ∪ (c ∩ a)) ≤ (c⊥ ∪ (c ∩ (a ∪ b)))
5	1, 4	bltr 138	. . 3 (c →1 a) ≤ (c⊥ ∪ (c ∩ (a ∪ b)))
6		df-i1 44	. . . 4 (c →1 b) = (c⊥ ∪ (c ∩ b))
7		leor 159	. . . . . 6 b ≤ (a ∪ b)
8	7	lelan 167	. . . . 5 (c ∩ b) ≤ (c ∩ (a ∪ b))
9	8	lelor 166	. . . 4 (c⊥ ∪ (c ∩ b)) ≤ (c⊥ ∪ (c ∩ (a ∪ b)))
10	6, 9	bltr 138	. . 3 (c →1 b) ≤ (c⊥ ∪ (c ∩ (a ∪ b)))
11	5, 10	lel2or 170	. 2 ((c →1 a) ∪ (c →1 b)) ≤ (c⊥ ∪ (c ∩ (a ∪ b)))
12		df-i1 44	. . 3 (c →1 (a ∪ b)) = (c⊥ ∪ (c ∩ (a ∪ b)))
13	12	ax-r1 35	. 2 (c⊥ ∪ (c ∩ (a ∪ b))) = (c →1 (a ∪ b))
14	11, 13	lbtr 139	1 ((c →1 a) ∪ (c →1 b)) ≤ (c →1 (a ∪ b))

This proof depends on syntax axioms:   ≤ wle 2  ^⊥ 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-t 41  df-f 42  df-i1 44  df-le1 130  df-le2 131

This theorem is used by:  orbile  843