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May 5, 2019 at 8:34 vote accept lightsodium
May 5, 2019 at 7:53 answer added Rajesh Shashi Kumar timeline score: 0
May 5, 2019 at 7:06 history edited lightsodium CC BY-SA 4.0
added 104 characters in body
May 5, 2019 at 6:49 comment added lightsodium (a'+b)*(bc'+bd') = a'bc'+a'bd'+bbc' + bbd' = a'bc' + a'bd' + bc' + bd' , do you mean this?
May 4, 2019 at 20:50 comment added JRE I'm saying that if you multiply things out to get b(cd)' , then you ought to have something with a'(some stuff) as well, and that'll be + b(cd)' . And, will that be simpler (or can it be simplified to something simpler) than your current result?
May 4, 2019 at 20:45 comment added lightsodium So you would recommend to factorise b ?: b ( a' + c'd')?
May 4, 2019 at 20:44 comment added JRE But then where's the rest of it?
May 4, 2019 at 20:42 comment added lightsodium How do you get bc' + bd' out of b(c*d)'? : By multplying the brackets out
May 4, 2019 at 20:36 comment added JRE How do you get bc' + bd' out of b(c*d)'? For that fact, how do you figure to get that b out of (a'+b)? Wouldn't you have to do something with the a'?
May 4, 2019 at 20:12 comment added lightsodium Is the expression to the circuit now right?
May 4, 2019 at 19:54 comment added lightsodium Now, there should be the right gate.
May 4, 2019 at 19:54 history edited lightsodium CC BY-SA 4.0
added 2 characters in body
May 4, 2019 at 19:53 comment added lightsodium Oh, sorry, I will edit sth. in the picture.
May 4, 2019 at 19:48 comment added JRE Does + really mean "NOR?"
May 4, 2019 at 19:41 history edited lightsodium CC BY-SA 4.0
added 60 characters in body; edited title
May 4, 2019 at 19:41 history edited JRE CC BY-SA 4.0
edited body; edited title
May 4, 2019 at 19:39 history asked lightsodium CC BY-SA 4.0