Answers
Feb 09, 2012 - 12:16 AM
The distributive property applies to both and and or operations. To summarize:
a | (b & c) === (a | b) & (a | c) /// distribution on |
a & (b | c) === (a & b) | (a & c) /// distribution on &
Your example requires multiple distributions. Algebraic substitution with a => x,y on the first example yields the following result:
(a=(x & y)) | (b & c) === (x&y | b) & (x&y | c)
=== (b | x)&(b | y)&(c | x)&(c | y)
expanding this thinking yields:
(abc)|(xyz) ===>>
(a|x)&(a|y)&(a|z) &
(b|x)&(b|y)&(b|z) &
(c|x)&(c|y)&(c|z) &
now if we want to distribute (i|j|k) on the result, we would then need to grind through each or term. Taking a
small snap-shop as follows:
(i|j|k)&(a|x) ===>>
(i|a)&(i|x)&(j|a)&(j|x)&(k|a)&(k|x)
Further expansion blows up pretty big.
a | (b & c) === (a | b) & (a | c) /// distribution on |
a & (b | c) === (a & b) | (a & c) /// distribution on &
Your example requires multiple distributions. Algebraic substitution with a => x,y on the first example yields the following result:
(a=(x & y)) | (b & c) === (x&y | b) & (x&y | c)
=== (b | x)&(b | y)&(c | x)&(c | y)
expanding this thinking yields:
(abc)|(xyz) ===>>
(a|x)&(a|y)&(a|z) &
(b|x)&(b|y)&(b|z) &
(c|x)&(c|y)&(c|z) &
now if we want to distribute (i|j|k) on the result, we would then need to grind through each or term. Taking a
small snap-shop as follows:
(i|j|k)&(a|x) ===>>
(i|a)&(i|x)&(j|a)&(j|x)&(k|a)&(k|x)
Further expansion blows up pretty big.
Mar 31, 2012 - 10:38 AM
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