搜尋自 英語 {1} 教師……
izzet
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
解答 · 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
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亞塞拜然語 (Azeri), 英語, 庫德語, 土耳其語
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