Theorem wql1 285

Description: The 2nd hypothesis is the first →₁ WQL axiom. We show it implies the WOM law.

Hypotheses

Ref	Expression
wql1.1	(a →1 b) = 1
wql1.2	((a ∪ c) →1 (b ∪ c)) = 1
wql1.3	c = b

Assertion

Ref	Expression
wql1	(a →2 b) = 1

Proof of Theorem wql1

Step	Hyp	Ref	Expression
1		df-i2 44	. 2 (a →2 b) = (b ∪ (a⊥ ∩ b⊥ ))
2		anor3 82	. . 3 (a⊥ ∩ b⊥ ) = (a ∪ b)⊥
3	2	lor 66	. 2 (b ∪ (a⊥ ∩ b⊥ )) = (b ∪ (a ∪ b)⊥ )
4		ax-a2 30	. . 3 (b ∪ (a ∪ b)⊥ ) = ((a ∪ b)⊥ ∪ b)
5		wql1.3	. . . . . . . . 9 c = b
6	5	lor 66	. . . . . . . 8 (b ∪ c) = (b ∪ b)
7		oridm 102	. . . . . . . 8 (b ∪ b) = b
8	6, 7	ax-r2 35	. . . . . . 7 (b ∪ c) = b
9	8	ud1lem0a 247	. . . . . 6 ((a ∪ c) →1 (b ∪ c)) = ((a ∪ c) →1 b)
10	9	ax-r1 34	. . . . 5 ((a ∪ c) →1 b) = ((a ∪ c) →1 (b ∪ c))
11	5	lor 66	. . . . . 6 (a ∪ c) = (a ∪ b)
12	11	ud1lem0b 248	. . . . 5 ((a ∪ c) →1 b) = ((a ∪ b) →1 b)
13		wql1.2	. . . . 5 ((a ∪ c) →1 (b ∪ c)) = 1
14	10, 12, 13	3tr2 61	. . . 4 ((a ∪ b) →1 b) = 1
15	14	wql1lem 279	. . 3 ((a ∪ b)⊥ ∪ b) = 1
16	4, 15	ax-r2 35	. 2 (b ∪ (a ∪ b)⊥ ) = 1
17	1, 3, 16	3tr 62	1 (a →2 b) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₁ wi1 13   →₂ wi2 14

This theorem was proved from axioms:  ax-a1 29  ax-a2 30  ax-a3 31  ax-a4 32  ax-a5 33  ax-r1 34  ax-r2 35  ax-r4 36  ax-r5 37

This theorem depends on definitions:  df-a 39  df-t 40  df-f 41  df-i1 43  df-i2 44  df-le1 122  df-le2 123

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