Theorem u1lem3 749

Description: Lemma for unified implication study. (Contributed by NM, 17-Dec-1997.)

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 44	. 2 (a →1 (b →1 a)) = (a⊥ ∪ (a ∩ (b →1 a)))
2		ancom 74	. . . . . . . 8 (a ∩ b) = (b ∩ a)
3		ancom 74	. . . . . . . 8 (a ∩ b⊥ ) = (b⊥ ∩ a)
4	2, 3	2or 72	. . . . . . 7 ((a ∩ b) ∪ (a ∩ b⊥ )) = ((b ∩ a) ∪ (b⊥ ∩ a))
5		u1lemab 610	. . . . . . . 8 ((b →1 a) ∩ a) = ((b ∩ a) ∪ (b⊥ ∩ a))
6	5	ax-r1 35	. . . . . . 7 ((b ∩ a) ∪ (b⊥ ∩ a)) = ((b →1 a) ∩ a)
7	4, 6	ax-r2 36	. . . . . 6 ((a ∩ b) ∪ (a ∩ b⊥ )) = ((b →1 a) ∩ a)
8		ancom 74	. . . . . 6 ((b →1 a) ∩ a) = (a ∩ (b →1 a))
9	7, 8	ax-r2 36	. . . . 5 ((a ∩ b) ∪ (a ∩ b⊥ )) = (a ∩ (b →1 a))
10	9	ax-r1 35	. . . 4 (a ∩ (b →1 a)) = ((a ∩ b) ∪ (a ∩ b⊥ ))
11	10	lor 70	. . 3 (a⊥ ∪ (a ∩ (b →1 a))) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
12		id 59	. . 3 (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ ))) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
13	11, 12	ax-r2 36	. 2 (a⊥ ∪ (a ∩ (b →1 a))) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))
14	1, 13	ax-r2 36	1 (a →1 (b →1 a)) = (a⊥ ∪ ((a ∩ b) ∪ (a ∩ b⊥ )))

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

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-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is used by:  u1lem4  757