Theorem ud2lem3 547

Description: Lemma for unified disjunction.

Assertion

Ref	Expression
ud2lem3	((a →2 b) →2 (a ∪ b)) = (a ∪ b)

Proof of Theorem ud2lem3

Step	Hyp	Ref	Expression
1		df-i2 44	. 2 ((a →2 b) →2 (a ∪ b)) = ((a ∪ b) ∪ ((a →2 b)⊥ ∩ (a ∪ b)⊥ ))
2		ud2lem0c 270	. . . . 5 (a →2 b)⊥ = (b⊥ ∩ (a ∪ b))
3	2	ran 71	. . . 4 ((a →2 b)⊥ ∩ (a ∪ b)⊥ ) = ((b⊥ ∩ (a ∪ b)) ∩ (a ∪ b)⊥ )
4	3	lor 66	. . 3 ((a ∪ b) ∪ ((a →2 b)⊥ ∩ (a ∪ b)⊥ )) = ((a ∪ b) ∪ ((b⊥ ∩ (a ∪ b)) ∩ (a ∪ b)⊥ ))
5		coman2 178	. . . . . 6 (b⊥ ∩ (a ∪ b)) C (a ∪ b)
6	5	comcom 435	. . . . 5 (a ∪ b) C (b⊥ ∩ (a ∪ b))
7		comid 179	. . . . . 6 (a ∪ b) C (a ∪ b)
8	7	comcom2 175	. . . . 5 (a ∪ b) C (a ∪ b)⊥
9	6, 8	fh3 453	. . . 4 ((a ∪ b) ∪ ((b⊥ ∩ (a ∪ b)) ∩ (a ∪ b)⊥ )) = (((a ∪ b) ∪ (b⊥ ∩ (a ∪ b))) ∩ ((a ∪ b) ∪ (a ∪ b)⊥ ))
10		ancom 68	. . . . . . 7 (b⊥ ∩ (a ∪ b)) = ((a ∪ b) ∩ b⊥ )
11	10	lor 66	. . . . . 6 ((a ∪ b) ∪ (b⊥ ∩ (a ∪ b))) = ((a ∪ b) ∪ ((a ∪ b) ∩ b⊥ ))
12		df-t 40	. . . . . . 7 1 = ((a ∪ b) ∪ (a ∪ b)⊥ )
13	12	ax-r1 34	. . . . . 6 ((a ∪ b) ∪ (a ∪ b)⊥ ) = 1
14	11, 13	2an 72	. . . . 5 (((a ∪ b) ∪ (b⊥ ∩ (a ∪ b))) ∩ ((a ∪ b) ∪ (a ∪ b)⊥ )) = (((a ∪ b) ∪ ((a ∪ b) ∩ b⊥ )) ∩ 1)
15		an1 98	. . . . . 6 (((a ∪ b) ∪ ((a ∪ b) ∩ b⊥ )) ∩ 1) = ((a ∪ b) ∪ ((a ∪ b) ∩ b⊥ ))
16		a5b 112	. . . . . 6 ((a ∪ b) ∪ ((a ∪ b) ∩ b⊥ )) = (a ∪ b)
17	15, 16	ax-r2 35	. . . . 5 (((a ∪ b) ∪ ((a ∪ b) ∩ b⊥ )) ∩ 1) = (a ∪ b)
18	14, 17	ax-r2 35	. . . 4 (((a ∪ b) ∪ (b⊥ ∩ (a ∪ b))) ∩ ((a ∪ b) ∪ (a ∪ b)⊥ )) = (a ∪ b)
19	9, 18	ax-r2 35	. . 3 ((a ∪ b) ∪ ((b⊥ ∩ (a ∪ b)) ∩ (a ∪ b)⊥ )) = (a ∪ b)
20	4, 19	ax-r2 35	. 2 ((a ∪ b) ∪ ((a →2 b)⊥ ∩ (a ∪ b)⊥ )) = (a ∪ b)
21	1, 20	ax-r2 35	1 ((a →2 b) →2 (a ∪ b)) = (a ∪ b)

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7  1wt 9   →₂ wi2 14

This theorem is referenced by:  ud2 578

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-i2 44  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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