Hello, Im trying to solve this with identities xz+z /z + zxy = z z ( x +/x + xy) <-- x +/x = 1 (identity) z (xy) = z so now what? Im trying to find identity to get rid xy but I dont know how to do it ..
Where are you getting xz+z /z + zxy? Be careful with your distribution. Once you distribute everything correctly, you'll see a common factor. EDIT: Just realized you're using "/" as a "not" symbol. I suggest you use an apostrophe after the variable instead of a / before it.
Sorry.. xz + z(x'+xy) = z <-- x' + xy = x' + y xz + z ( x' + y ) = z xz + zx' + zy = z < -- distributive z ( x + x' + y ) = z <- x + x' = 1 z ( 1 + y ) = z <-- Im stuck! what can I do ?
Code ( (Unknown Language)): x' + xy = x' + y I've never seen this identity, if that was taught as one, great, if not, I think you need to expand this out...
but Im still stuck because you can see I carry out xz + z ( x' + y ) = z xz + zx' + zy = z < -- I expanded everything, and then the common factor is z and I got.. z ( x + x' + y ) = z <- since I have another x + x' = 1 according to my teacher A + A' = 1 z ( 1 + y ) = z <-- now Im not sure, I found another A + 1 = 1 ? is that can work 1 + y what do you think?
If anything is "OR"ed with a 1, the outcome is 1. So z (1 + y) is the same as z (1). What is Z AND 1?
Now I have another that Im stuck sorry x'+y' + xyz'= x'+y'+ z' No idea how to do it .. the only think I can find x' y' could be xy'' ( xy are not) but there are not common factors or something in this equation that can lead me to do something
Remember the rule A' + AB' is A' + B'. If you rearrange a bit, you will see that this can help you twice. Matt
Sorry I dont know that rule, but I have this one with only one not .. A + A'B, but iff I have the whole term together because is XYZ' and the rule says A' + AB' ? is that applies?
Hmm, I was thinking it worked with two NOTs just as it would with only 1. I could be wrong though. I haven't done this in a while Perhaps I should let someone else jump in here, just in case my memory is failing me....
Let: A = x B = y' EDIT: formatted, per WBahn's advice... Code ( (Unknown Language)): x' + xy = x' + y ^ ^^ ^ ^ | || | | A' A B' A' B'
Try putting it in a CODE block. You probably will still see it as proportional spacing while you are typing, but it will show up as monospaced when posted (or previewed).
This is confusing because it isn't an identity. It is valid, but you only see why when you go through the steps... A + A'B : Expand A A + AB + A'B : Factor B' A + B(A + A') : Using Identity A + A' = 1 A + B(1) : Using identity A(1) = A => A + A'B = A + B So, if we were to use another form of B, namely B': A + A'B' : Expand A A + AB' + A'B' : Factor B' A + B'(A + A') : Using Identity A + A' = 1 A + B'(1) : Using identity A(1) = A A + B' => A + A'B' = A + B'