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 v c) ->1 (b v 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 v (a_|_ ^ b_|_))
2		anor3 82	. . 3 (a_|_ ^ b_|_) = (a v b)_|_
3	2	lor 66	. 2 (b v (a_|_ ^ b_|_)) = (b v (a v b)_|_)
4		ax-a2 30	. . 3 (b v (a v b)_|_) = ((a v b)_|_ v b)
5		wql1.3	. . . . . . . . 9 c = b
6	5	lor 66	. . . . . . . 8 (b v c) = (b v b)
7		oridm 102	. . . . . . . 8 (b v b) = b
8	6, 7	ax-r2 35	. . . . . . 7 (b v c) = b
9	8	ud1lem0a 247	. . . . . 6 ((a v c) ->1 (b v c)) = ((a v c) ->1 b)
10	9	ax-r1 34	. . . . 5 ((a v c) ->1 b) = ((a v c) ->1 (b v c))
11	5	lor 66	. . . . . 6 (a v c) = (a v b)
12	11	ud1lem0b 248	. . . . 5 ((a v c) ->1 b) = ((a v b) ->1 b)
13		wql1.2	. . . . 5 ((a v c) ->1 (b v c)) = 1
14	10, 12, 13	3tr2 61	. . . 4 ((a v b) ->1 b) = 1
15	14	wql1lem 279	. . 3 ((a v b)_|_ v b) = 1
16	4, 15	ax-r2 35	. 2 (b v (a v b)_|_) = 1
17	1, 3, 16	3tr 62	1 (a ->2 b) = 1

Syntax hints:   = wb 1  _|_wn 4   v 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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