Question

Logic table (! in front means not):

F= (A!B+!AB)(B+C)+(AB+!AB)(B+C)

I need some help with the truth table for this one

A	!A	B	!B	C	!C	F
0	1	0	1	0	1
0	1	0	1	1	0
0	1	1	0	0	1
0	1	1	0	1	0
1	0	0	1	0	1
1	0	0	1	1	0
1	0	1	0	0	1
1	0	1	0	1	0

Answer 1

Sol Given that, F= CAJB+AB)(B+C) + b(4!B+JAB)! (B+C) Truth Table O 0 - - O سه 0 - O -- LA AB d F=(113 HAB ) ) (Bte) +(als Hab)) (bts! F 0 - (10 Hooto) +(1.0+(00/coto) 3 (0+1).0 total (o)! = 0+0=0 0 = (1.0+(0:0)')(0+1)+(1+0+00:0)')'(0+1)' (0+1).1+(out'c'=1to = 1 ||= (1-1 +(0-1)}(140) F(IIH( 13 ) Ho) | =(1+0-1 Hiti! I'=170 = 1 - (0.1+(-1/'>0+1)+(!1+(ov)'CHU) = (1+1) (1) titly (U' =1 to = || 1 - (0.0 +(10)) 0+0)+(0-0 +(10) co tos' (0+)(0) +(0+1!0' = 0+0 -0 0 = (0-0 + (0))(0+)+(00+1000+1)' = (0+1)(1+cott' (u'=1+0=1 (0.1+(11)') (140) + (0.1+(1:1)')'(140)' - (0+0) (1)+10+0)' () = 0+0=0 1 0 0 1 0 (0.1+(113')(1+1)+ (0-1461)!) CI+1)! =(0+0).1 tcoto)' Wa =oto=0 0 0 - - O 1 - - O - - NOTE 0.0 = 0 6.1 -0 1.0 - 0 1.1 = oto - o ofl = 1 Ito =1 IH1 IT o'zi 1=0