Simplify:
(x * y) + (~x * z) + (y * z)
I'm using ~ to represent NOT, * to represent AND, + to represent OR
I did:
(x * y) + (~x * z) + (y * z) = (x * y) + z(~x * y)
But I can't get any further from here
A boolean algebra calculator online says the simplification to the original expression is : (x * y) + (~x * z) but provides no steps
Another boolean algebra equivalence tester says
(x * y) + (~x * z) + (y * z) is logically equivalent to (x * y) + (~x * z)
Help simplifying (x * y) + (~x * z) + (y * z) would be greatly appreciated.
(x * y) + (~x * z) + (y * z)
(x * y) +z(~x *y)
y+z
How did you get y + z from (x * y) +z(~x *y)?
typo using your symbols. I drew a logic tree.
If x, true=1 if y
If not x, true if y and z
so, true if y and (x or z) Put that into your symbols.
To simplify the expression (x * y) + (~x * z) + (y * z), we can apply the distributive property to eliminate the common terms and simplify further.
Let's start by distributing the terms (~x * z) and (y * z):
(x * y) + (~x * z) + (y * z)
= (x * y) + z(~x + y)
Now, we can see that (~x + y) is an expression that cannot be simplified further because it contains both the complement (~) and the OR (+) operation. So, we leave it as it is.
Therefore, the simplified expression is:
(x * y) + z(~x + y)