I get that the expression is equivalent A alone.
Do a truth table on the final expression: .
You might think of it as sets: .
Simplify this Boolean Expression
A & ~((~A) & (~B)).
According to the truth table I did it should be equivalent to ~B but I can't get the boolean algebra to work out to that.
A & ~((~A) & (~B))
= A & (~~A or ~~B) Demorgan's law
= A & (A or B) Double negation
= (A & A) or (A & B) Distributive Property
And this is where I get stuck. If A&A = just A then you get into an infinate loop of distrubutive properties and never reach your goal.
Hello, algebrapro18!
Simplify this Boolean expression: .
According to the truth table I did,
it should be equivalent to . . . . no, it's equivalent to
. .
. .
. . . . . . All this is correct!
And this is where I get stuck.
If , then you get into an infinite loop
of distrubutive properties and never reach your goal. . . . . This is true
There should be (or we must establish) a theorem
. . which states: .