Using Boolean Algebra to Prove Equations

[Hypnos_16]

Homework Statement

There should be lines of some values to imply the "Not" form of them, however to make it easier, i'll just use the ¬ Symbol

(a) Let x, y be elements of a Boolean algebra. Prove from the axioms that (x · y) + x = x.

(b) Prove from the axioms of Boolean algebra that x · ¬( y · ¬x + ¬y ) = (x + x) · (y + 0). You can use DeMorgan’s and other identities we already derived in class.

(c) In the proof of completeness of Boolean algebra, we showed how to convert every formula to its “canonical DNF”: a normal form corresponding to the DNF obtained from the truth table without any simplifications.
Describe the normal form for the formula ¬( y · ¬x + ¬y )

Homework Equations

x + y = y + x
x · y = y · x
(x + y) + z = x + (y + z)
(x · y) · z = x · (y · z)
x · (y + z) = x · y + x · z
x + y · z = (x + y) · (x + z)
x + 0 = x
x · 1 = x
x + ¬x = 1
x · ¬x = 0
0 ≠ 1

The Attempt at a Solution

Part a, i don't even know how to start it. There doesn't seem to be anything i can do to it.

Part b, i have an attempt for
x * ¬(y * ¬x + ¬y) = (x + x) * (y + 0)
x * (¬y + x * y) = (x + x) * (y + 0)
x * (¬y + y * y + x) = (x + x) * (y + 0)
x * (1 * y + x) = (x + x) * (y + 0)
x * (x + 1) * (y + 1) = (x + x) * (y + 0)
x * x + x * 1 * (y + 1) = (x + x) * (y + 0)
0 + x * (y + 1) = (x + x) * (y + 0)
0 + (x * y) + (x * 1) = (x + x) * (y + 0)
0 + (x * y) + (x) = (x + x) * (y + 0)

but i get here and get stuck.

and part c, much like part a, i don't even know how to start it.

 

[ehild]

a) You did not include into the relevant equations but you certainly know that x+1=1.
There are two more axioms: 1*x=x and the distributive law: a*(b+c)=a*b+a*c.
Write y*x+x in the form y*x+1*x, apply the reverse of the distributive law, (factor out x) ...

b)

¬(y * ¬x + ¬y) ≠(¬y + x * y).

The correct application of de Morgan rule is :

¬((y * ¬x) + ¬y) =¬(y * ¬x)*y=(¬y+x)*y=...

c)Set up the truth table of the expression. Collect what product of x and y result 1 and add them. For example, if you get 1 when x= 0 and y =1 and also when x=1 and y = 1, then the canonical form is ¬x*y + x*y.

ehild