Theorem u1lemab 610

Description: Lemma for Sasaki implication study. Equation 4.10 of [MegPav2000] p. 23. This is the second part of the equation. (Contributed by NM, 14-Dec-1997.)

Assertion

Ref	Expression
u1lemab	((a →1 b) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))

Proof of Theorem u1lemab

Step	Hyp	Ref	Expression
1		df-i1 44	. . 3 (a →1 b) = (a⊥ ∪ (a ∩ b))
2	1	ran 78	. 2 ((a →1 b) ∩ b) = ((a⊥ ∪ (a ∩ b)) ∩ b)
3		ax-a2 31	. . . . 5 (a⊥ ∪ (a ∩ b)) = ((a ∩ b) ∪ a⊥ )
4	3	ran 78	. . . 4 ((a⊥ ∪ (a ∩ b)) ∩ b) = (((a ∩ b) ∪ a⊥ ) ∩ b)
5		coman2 186	. . . . 5 (a ∩ b) C b
6		coman1 185	. . . . . 6 (a ∩ b) C a
7	6	comcom2 183	. . . . 5 (a ∩ b) C a⊥
8	5, 7	fh2r 474	. . . 4 (((a ∩ b) ∪ a⊥ ) ∩ b) = (((a ∩ b) ∩ b) ∪ (a⊥ ∩ b))
9	4, 8	ax-r2 36	. . 3 ((a⊥ ∪ (a ∩ b)) ∩ b) = (((a ∩ b) ∩ b) ∪ (a⊥ ∩ b))
10		anass 76	. . . . 5 ((a ∩ b) ∩ b) = (a ∩ (b ∩ b))
11		anidm 111	. . . . . 6 (b ∩ b) = b
12	11	lan 77	. . . . 5 (a ∩ (b ∩ b)) = (a ∩ b)
13	10, 12	ax-r2 36	. . . 4 ((a ∩ b) ∩ b) = (a ∩ b)
14	13	ax-r5 38	. . 3 (((a ∩ b) ∩ b) ∪ (a⊥ ∩ b)) = ((a ∩ b) ∪ (a⊥ ∩ b))
15	9, 14	ax-r2 36	. 2 ((a⊥ ∪ (a ∩ b)) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))
16	2, 15	ax-r2 36	1 ((a →1 b) ∩ b) = ((a ∩ b) ∪ (a⊥ ∩ b))

Syntax hints:   = wb 1  ^⊥ wn 4   ∪ wo 6   ∩ wa 7   →₁ wi1 12

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-i1 44  df-le1 130  df-le2 131  df-c1 132  df-c2 133

This theorem is referenced by:  u1lemnonb  675  i1com  708  u1lem3  749  u1lem11  780  sadm3  838  negantlem2  849  negantlem3  850  negantlem10  861  neg3antlem1  864  oa4to4u  973  oa3-6lem  980  oa3-u1lem  985  oa3-u2lem  986  oa3-1to5  993  lem4.6.2e2  1083