A.not(B)+AC

nepaliraja · Oct 16, 2009 7:51 AM

abc + aB + aBc ac(b+B) + aB ac x 0 + aB aB B = your Bold B Not sure if this is correct. Last edited: 16-Oct-09 10:49 AM

yellow · Oct 16, 2009 10:47 AM

I have made little bit clear The confusion here is, it seems it is giving to different answers A.B.C +A.Bnot Another is A.C+A.Bnot I am little bit confused here. Does it has two solution?  

timer · Oct 16, 2009 11:12 AM

Strictly following the precedence i.e (. before +) A.B.C  + A. B (not A . not C) (De morgans law) A.B.C + A.B.notA.notC A.B.C + (A.notA).B.notC A and not A = False ABC + False so if any of the A or B or C is false the result is False and for the expression to be true they all have to be True I am assuming that the bnot.(A+C) = b.not(A+C) Please consult the solution below, the above is solved on wrong assumption :D Last edited: 16-Oct-09 11:22 AM Last edited: 16-Oct-09 11:40 AM

dynamite · Oct 16, 2009 11:15 AM

assuming bnot = not b a.b.c + a.notb.a + a.notb.c (distributive)   a.b.c + a.notb + a.notbc ( a.a = a) a.c.b   + a.c.notb  +a.notb (re-arranging) ac(b+ notb) + a.notb ( distributive) b+ notb = true therefore ac + a.notb Last edited: 16-Oct-09 11:36 AM

dynamite · Oct 16, 2009 11:31 AM