Theorem u3lem8 783

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 508	. . . 4 a C (a →3 (a⊥ →3 b))
2	1	comcom3 454	. . 3 a⊥ C (a →3 (a⊥ →3 b))
3	2	u3lemc4 703	. 2 (a⊥ →3 (a →3 (a⊥ →3 b))) = (a⊥ ⊥ ∪ (a →3 (a⊥ →3 b)))
4		ax-a1 30	. . . . 5 a = a⊥ ⊥
5	4	ax-r1 35	. . . 4 a⊥ ⊥ = a
6		u3lem7 774	. . . 4 (a →3 (a⊥ →3 b)) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
7	5, 6	2or 72	. . 3 (a⊥ ⊥ ∪ (a →3 (a⊥ →3 b))) = (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))))
8		ax-a3 32	. . . . 5 ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))))
9	8	ax-r1 35	. . . 4 (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))) = ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
10		ax-a2 31	. . . . 5 ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ (a ∪ a⊥ ))
11		df-t 41	. . . . . . . 8 1 = (a ∪ a⊥ )
12	11	ax-r1 35	. . . . . . 7 (a ∪ a⊥ ) = 1
13	12	lor 70	. . . . . 6 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ (a ∪ a⊥ )) = (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ 1)
14		or1 104	. . . . . 6 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ 1) = 1
15	13, 14	ax-r2 36	. . . . 5 (((a ∩ b) ∪ (a ∩ b⊥ )) ∪ (a ∪ a⊥ )) = 1
16	10, 15	ax-r2 36	. . . 4 ((a ∪ a⊥ ) ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = 1
17	9, 16	ax-r2 36	. . 3 (a ∪ (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))) = 1
18	7, 17	ax-r2 36	. 2 (a⊥ ⊥ ∪ (a →3 (a⊥ →3 b))) = 1
19	3, 18	ax-r2 36	1 (a⊥ →3 (a →3 (a⊥ →3 b))) = 1

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₃ wi3 14

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a4 33  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38  ax-r3 439

This theorem depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i3 46  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by: (None)