Theorem anandir 115

Description: Distribution of conjunction over conjunction. (Contributed by NM, 27-Aug-1997.)

Assertion

Ref	Expression
anandir	((a ∩ b) ∩ c) = ((a ∩ c) ∩ (b ∩ c))

Proof of Theorem anandir

Step	Hyp	Ref	Expression
1		anidm 111	. . . 4 (c ∩ c) = c
2	1	ax-r1 35	. . 3 c = (c ∩ c)
3	2	lan 77	. 2 ((a ∩ b) ∩ c) = ((a ∩ b) ∩ (c ∩ c))
4		an4 86	. 2 ((a ∩ b) ∩ (c ∩ c)) = ((a ∩ c) ∩ (b ∩ c))
5	3, 4	ax-r2 36	1 ((a ∩ b) ∩ c) = ((a ∩ c) ∩ (b ∩ c))

Syntax hints:   = wb 1   ∩ wa 7

This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-a5 34  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38

This theorem depends on definitions:  df-a 40  df-t 41  df-f 42

This theorem is referenced by:  leran  153  ka4lemo  228  wr5-2v  366  wleran  394  ska4  433  i3orlem5  556  ud2lem1  563  mlaoml  833  comanblem2  871  e2astlem1  895  oath1  1004  4oath1  1041  lem3.3.6  1056