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Who can solve this of math question? (logical question)
[p ^ ( p' v q ) ]' v q
Which is the most simple way?
a) 1 b) 0 c) p v q d) p ^ q e) q.
. answer ; [ p ^ (p' v q )]' v q = [ p' v (p' v q)'] v q , ( De Morgan ) .....................................
= [ p' v ((p')' ^ q') ] v q ...............................................................................................
= [ p' v (p ^ q' ) ] v q , (distributive property) ................................................................
= [(p' v p) ^ ( p' v q')] v q ...........................................................................................
= [ 1 ^ ( p' v q' )] v q .................................................................................................
= (p' v q' ) v q , (coupling) .........................................................................................
= p' v ( q' v q ) ........................................................................................................
= p' v 1 .................................................................................................................
== 1 // .................................................................................................................
2012年6月7日 05:14
回答 · 8
1
Imagine that p and q are circles that overlap. It's useful to draw a series of pictures and darken the parts we want.
The innermost parenthesis consists of everything that is either not p or is q. So it consists of everything except that part of circle p that doesn't overlap with q.
The next innermost parenthesis consists of what we've just determined that is also in p. That's just the overlapping parts of p and q.
That is inversed, so we now have everything except the overlapping parts of p and q.
Finally we want everything that is either outside the overlapping parts or is in q. That's everything.
So your answer is a)
2012年6月9日
answer ; [ p ^ (p' v q )]' v q = [ p' v (p' v q)'] v q , ( De Morgan ) .........................................................................................
= [ p' v ((p')' ^ q') ] v q .........................................................................................
= [ p' v (p ^ q' ) ] v q , (distributive property) .....................................
= [(p' v p) ^ ( p' v q')] v q .........................................................................................
= [ 1 ^ ( p' v q' )] v q .........................................................................................
= (p' v q' ) v q , (coupling) .........................................................................................
= p' v ( q' v q ) .........................................................................................
= p' v 1 .........................................................................................
== 1 // .........................................................................................
2012年6月9日
answer ;
[ p ^ (p' v q )]' v q = [ p' v (p' v q)'] v q , ( De Morgan )
= [ p' v ((p')' ^ q') ] v q
= [ p' v (p ^ q' ) ] v q , (distributive property)
= [(p' v p) ^ ( p' v q')] v q
= [ 1 ^ ( p' v q' )] v q
= (p' v q' ) v q , (coupling)
= p' v ( q' v q )
= p' v 1
== 1 //
2012年6月9日
@Liz : it is math question :) but logical...
.
.
@dnurfiani : ~p and p' same just different write :)
2012年6月7日
have no idea... what's that??
2012年6月7日
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izzet
语言技能
阿塞拜疆语 (Azeri), 英语, 库尔德语, 土耳其语
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英语
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