## Answers

Feb 09, 2012 - 01:16 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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