Theorem u1lem3 731

Description: Lemma for unified implication study.

Assertion

Ref	Expression
u1lem3	(a →1 (b →1 a)) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))

Proof of Theorem u1lem3

Step	Hyp	Ref	Expression
1		df-i1 43	. 2 (a →1 (b →1 a)) = (a⊥ ∪ (a ∩ (b →1 a)))
2		ancom 68	. . . . . . . 8 (a ∩ b) = (b ∩ a)
3		ancom 68	. . . . . . . 8 (a ∩ b⊥ ) = (b⊥ ∩ a)
4	2, 3	2or 67	. . . . . . 7 ((a ∩ b) ∪ (a ∩ b⊥ )) = ((b ∩ a) ∪ (b⊥ ∩ a))
5		u1lemab 592	. . . . . . . 8 ((b →1 a) ∩ a) = ((b ∩ a) ∪ (b⊥ ∩ a))
6	5	ax-r1 34	. . . . . . 7 ((b ∩ a) ∪ (b⊥ ∩ a)) = ((b →1 a) ∩ a)
7	4, 6	ax-r2 35	. . . . . 6 ((a ∩ b) ∪ (a ∩ b⊥ )) = ((b →1 a) ∩ a)
8		ancom 68	. . . . . 6 ((b →1 a) ∩ a) = (a ∩ (b →1 a))
9	7, 8	ax-r2 35	. . . . 5 ((a ∩ b) ∪ (a ∩ b⊥ )) = (a ∩ (b →1 a))
10	9	ax-r1 34	. . . 4 (a ∩ (b →1 a)) = ((a ∩ b) ∪ (a ∩ b⊥ ))
11	10	lor 66	. . 3 (a⊥ ∪ (a ∩ (b →1 a))) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
12		id 58	. . 3 (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
13	11, 12	ax-r2 35	. 2 (a⊥ ∪ (a ∩ (b →1 a))) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
14	1, 13	ax-r2 35	1 (a →1 (b →1 a)) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 13

This theorem is referenced by:  u1lem4 739

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-i1 43  df-le1 122  df-le2 123  df-c1 124  df-c2 125

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