Theorem wql2lem3 282

Description: Lemma for ->2 WQL axiom.

Hypothesis

Ref	Expression
wql2lem3.1	(a ->2 b) = 1

Assertion

Ref	Expression
wql2lem3	((a ^ b_|_) ->2 a_|_) = 1

Proof of Theorem wql2lem3

Step	Hyp	Ref	Expression
1		df-i2 44	. 2 ((a ^ b_|_) ->2 a_|_) = (a_|_ v ((a ^ b_|_)_|_ ^ a_|__|_))
2		oran2 84	. . . . . 6 (a_|_ v b) = (a ^ b_|_)_|_
3	2	ax-r1 34	. . . . 5 (a ^ b_|_)_|_ = (a_|_ v b)
4	3	ran 71	. . . 4 ((a ^ b_|_)_|_ ^ a_|__|_) = ((a_|_ v b) ^ a_|__|_)
5		ancom 68	. . . 4 ((a_|_ v b) ^ a_|__|_) = (a_|__|_ ^ (a_|_ v b))
6	4, 5	ax-r2 35	. . 3 ((a ^ b_|_)_|_ ^ a_|__|_) = (a_|__|_ ^ (a_|_ v b))
7	6	lor 66	. 2 (a_|_ v ((a ^ b_|_)_|_ ^ a_|__|_)) = (a_|_ v (a_|__|_ ^ (a_|_ v b)))
8		wql2lem3.1	. . . 4 (a ->2 b) = 1
9	8	wql2lem 280	. . 3 (a_|_ v b) = 1
10		omlem2 120	. . 3 ((a_|_ v b)_|_ v (a_|_ v (a_|__|_ ^ (a_|_ v b)))) = 1
11	9, 10	skr0 234	. 2 (a_|_ v (a_|__|_ ^ (a_|_ v b))) = 1
12	1, 7, 11	3tr 62	1 ((a ^ b_|_) ->2 a_|_) = 1

Syntax hints:   = wb 1  _|_wn 4   v wo 6   ^ wa 7  1wt 9   ->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-i2 44  df-le1 122  df-le2 123

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