Theorem u3lem14a 791

Description: Lemma for unified implication study. (Contributed by NM, 19-Jan-1998.)

Assertion

Ref	Expression
u3lem14a	(a →3 ((b →3 a⊥ ) →3 b⊥ )) = (a →3 (b →3 a))

Proof of Theorem u3lem14a

Step	Hyp	Ref	Expression
1		u3lem13b 790	. . 3 ((b →3 a⊥ ) →3 b⊥ ) = (b →1 a)
2	1	ud3lem0a 260	. 2 (a →3 ((b →3 a⊥ ) →3 b⊥ )) = (a →3 (b →1 a))
3		df-i3 46	. . 3 (a →3 (b →1 a)) = (((a⊥ ∩ (b →1 a)) ∪ (a⊥ ∩ (b →1 a)⊥ )) ∪ (a ∩ (a⊥ ∪ (b →1 a))))
4		ancom 74	. . . . . . . 8 (a⊥ ∩ (b →1 a)) = ((b →1 a) ∩ a⊥ )
5		u1lemanb 615	. . . . . . . 8 ((b →1 a) ∩ a⊥ ) = (b⊥ ∩ a⊥ )
6	4, 5	ax-r2 36	. . . . . . 7 (a⊥ ∩ (b →1 a)) = (b⊥ ∩ a⊥ )
7		ancom 74	. . . . . . . 8 (a⊥ ∩ (b →1 a)⊥ ) = ((b →1 a)⊥ ∩ a⊥ )
8		u1lemnanb 655	. . . . . . . 8 ((b →1 a)⊥ ∩ a⊥ ) = (b ∩ a⊥ )
9	7, 8	ax-r2 36	. . . . . . 7 (a⊥ ∩ (b →1 a)⊥ ) = (b ∩ a⊥ )
10	6, 9	2or 72	. . . . . 6 ((a⊥ ∩ (b →1 a)) ∪ (a⊥ ∩ (b →1 a)⊥ )) = ((b⊥ ∩ a⊥ ) ∪ (b ∩ a⊥ ))
11		ax-a2 31	. . . . . . 7 ((b⊥ ∩ a⊥ ) ∪ (b ∩ a⊥ )) = ((b ∩ a⊥ ) ∪ (b⊥ ∩ a⊥ ))
12		ancom 74	. . . . . . . 8 (b ∩ a⊥ ) = (a⊥ ∩ b)
13		ancom 74	. . . . . . . 8 (b⊥ ∩ a⊥ ) = (a⊥ ∩ b⊥ )
14	12, 13	2or 72	. . . . . . 7 ((b ∩ a⊥ ) ∪ (b⊥ ∩ a⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
15	11, 14	ax-r2 36	. . . . . 6 ((b⊥ ∩ a⊥ ) ∪ (b ∩ a⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
16	10, 15	ax-r2 36	. . . . 5 ((a⊥ ∩ (b →1 a)) ∪ (a⊥ ∩ (b →1 a)⊥ )) = ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))
17		ax-a2 31	. . . . . . . 8 (a⊥ ∪ (b →1 a)) = ((b →1 a) ∪ a⊥ )
18		u1lemonb 635	. . . . . . . 8 ((b →1 a) ∪ a⊥ ) = 1
19	17, 18	ax-r2 36	. . . . . . 7 (a⊥ ∪ (b →1 a)) = 1
20	19	lan 77	. . . . . 6 (a ∩ (a⊥ ∪ (b →1 a))) = (a ∩ 1)
21		an1 106	. . . . . 6 (a ∩ 1) = a
22	20, 21	ax-r2 36	. . . . 5 (a ∩ (a⊥ ∪ (b →1 a))) = a
23	16, 22	2or 72	. . . 4 (((a⊥ ∩ (b →1 a)) ∪ (a⊥ ∩ (b →1 a)⊥ )) ∪ (a ∩ (a⊥ ∪ (b →1 a)))) = (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ a)
24		ax-a2 31	. . . . 5 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ a) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
25		u3lem3 751	. . . . . . 7 (a →3 (b →3 a)) = (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )))
26	25	ax-r1 35	. . . . . 6 (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (a →3 (b →3 a))
27		id 59	. . . . . 6 (a →3 (b →3 a)) = (a →3 (b →3 a))
28	26, 27	ax-r2 36	. . . . 5 (a ∪ ((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ ))) = (a →3 (b →3 a))
29	24, 28	ax-r2 36	. . . 4 (((a⊥ ∩ b) ∪ (a⊥ ∩ b⊥ )) ∪ a) = (a →3 (b →3 a))
30	23, 29	ax-r2 36	. . 3 (((a⊥ ∩ (b →1 a)) ∪ (a⊥ ∩ (b →1 a)⊥ )) ∪ (a ∩ (a⊥ ∪ (b →1 a)))) = (a →3 (b →3 a))
31	3, 30	ax-r2 36	. 2 (a →3 (b →1 a)) = (a →3 (b →3 a))
32	2, 31	ax-r2 36	1 (a →3 ((b →3 a⊥ ) →3 b⊥ )) = (a →3 (b →3 a))

This proof depends on syntax axioms:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 8   →₁ wi1 12   →₃ wi3 14

This proof depends on 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 proof depends on definitions:  df-b 39  df-a 40  df-t 41  df-f 42  df-i1 44  df-i3 46  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  u3lem14aa  792