Theorem u3lem8 765

Description: Lemma for unified implication study.

Assertion

Ref	Expression
u3lem8	(a⊥ →3 (a →3 (a⊥ →3 b))) = 1

Proof of Theorem u3lem8

Step	Hyp	Ref	Expression
1		comi31 490	. . . 4 a C (a →3 (a⊥ →3 b))
2	1	comcom3 436	. . 3 a⊥ C (a →3 (a⊥ →3 b))
3	2	u3lemc4 685	. 2 (a⊥ →3 (a →3 (a⊥ →3 b))) = (a⊥ ⊥ ∪ (a →3 (a⊥ →3 b)))
4		ax-a1 29	. . . . 5 a = a⊥ ⊥
5	4	ax-r1 34	. . . 4 a⊥ ⊥ = a
6		u3lem7 756	. . . 4 (a →3 (a⊥ →3 b)) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
7	5, 6	2or 67	. . 3 (a⊥ ⊥ ∪ (a →3 (a⊥ →3 b))) = (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))))
8		ax-a3 31	. . . . 5 ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))))
9	8	ax-r1 34	. . . 4 (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))) = ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
10		ax-a2 30	. . . . 5 ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ (a ∪ a⊥ ))
11		df-t 40	. . . . . . . 8 1 = (a ∪ a⊥ )
12	11	ax-r1 34	. . . . . . 7 (a ∪ a⊥ ) = 1
13	12	lor 66	. . . . . 6 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ (a ∪ a⊥ )) = (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ 1)
14		or1 96	. . . . . 6 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ 1) = 1
15	13, 14	ax-r2 35	. . . . 5 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ (a ∪ a⊥ )) = 1
16	10, 15	ax-r2 35	. . . 4 ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = 1
17	9, 16	ax-r2 35	. . 3 (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))) = 1
18	7, 17	ax-r2 35	. 2 (a⊥ ⊥ ∪ (a →3 (a⊥ →3 b))) = 1
19	3, 18	ax-r2 35	1 (a⊥ →3 (a →3 (a⊥ →3 b))) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₃ wi3 15

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  ax-r3 421

This theorem depends on definitions:  df-b 38  df-a 39  df-t 40  df-f 41  df-i3 45  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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