Simplify the Boolean expression " (x+y).(x+z) " .
I think answer is " x+y.z " But i don't know how t get that.
You should use the De Morgan Law (A+B)=(A'.B'). It works this way:
(X+Y)=X'.Y' and (X+Z)=X'.Z'
By commutativity: (X+Y).(X+Z)=(X'.Y').(X'.Z')=X'.Y'.X'.Z'=X'.X'.Y'.Z'
By idempotence: X'.X'=X'
Then: X'.X'.Y'.Z'=X'.Y'.Z'=X'.(Y'.Z')
Calling: Y'.Z'=W
Then: X'.(Y'.Z')=X'.W'
By De Morgan: X'.W'=(X+W) (I)
Negating the affirmation: W'=Y'.Z' then W=(Y'.Z')'=Y'+Z'=Y.Z (II)
By (I) and (II): (X+Y).(X+Z)=X+(Y.Z)=X+Y.Z
(x+y)(x+z)
= xx + xz + yx + yz
= x + xz + yx + yz (since xx = x eg 0.0 = 0 , 1.1 = 1)
= x(1 + z + y) + yz
= x(1 + y) +yz (since 1 + z = 1 e.g 1+0 = 1 or 1+1 = 1)
= x(1) + yz (since 1 +y =1 as explained above)
= x + yz
(x+y)(x+z) -Distribute-> xx+xy+xz+yz -x.x=x-> x+xy+xz+yz -> x+x(y+z)+yz -x=x.1-> x.1+x(y+z)+yz -> x(1+(y+z))+yz -1+(y+z)=1-> x+yz
Here is a much simpler solution by using idempotent(xx = x) and absorption(x+xy = x) laws.
(x+y)(x+z) = xx+xz+xy+yz = x+yz
Related
I am solving a fourth order equation in matlab using the solve function.My script looks like this:
syms m M I L Bp Bc g x
m = 0.127
M = 1.206
I = 0.001
L = 0.178
Bc = 5.4
Bp = 0.002
g = 9.8
eqn = ((m + M)*(I + m*L^2) - m^2*L^2)*x^4 + ((m + M)*Bp + (I + m*L^2)*Bc)*x^3 + ((m + M)*m*g*L + Bc*Bp)*x^2 + m*g*L*Bc*x == 0
S = solve(eqn, x)
In the answer, I should get 4 roots, but instead I get such strange expressions:
S =
0
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 1)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 2)
root(z^3 + (34351166180215288*z^2)/7131724328013535 + (352922208800606144*z)/7131724328013535 + 1379250971773894912/7131724328013535, z, 3)
The first root, which is 0, is displayed clearly. Is it possible to make the other three roots appear as numbers as well? I looked for something about this in the documentation for the solve function, but did not find it.
I have to restrict it with the folowings:
P(-1) = f(-1), P(0)=f(0), P(1)=f(1), P'(1)=f'(1)
Let the polynomial be
ax³ + bx² + cx + d
By the given equations,
- a + b - c + d = f(-1)
d = f(0)
a + b + c + d = f(1)
3a +2b + c = f'(1)
You should be able to solve.
Using DeMorgan's theorem show that:
a. (A + B)'(A' +B)' = 0
b. A + A'B + A'B' = 1
Boolean expression F = x'y + xyz':
Derive an algebraic expression for the complement F'
Show that F·F' = 0
Show that F + F' = 1
Please Help me
Assuming you know how DeMorgan's law works and you understand the basics of AND, OR, NOT operations:
1.a) (A + B)'(A' + B)' = A'B'(A')'B' = A'B'AB' = A'AB'B' = A'AB' = 0 B' = 0.
I used two facts here that hold for any boolean variable A:
AA' = 0 (when A = 0, A' = 1 and when A = 1, A' = 0 so (AA') has to be 0)
0A = 0 (0 AND anything has to be 0)
1.b) A + A'B + A'B' = A + A'(B + B') = A + A' = 1.
I used the following two facts that hold for any boolean variables A, B and C:
AB + AC = A(B + C) - just like you would do with numeric variables and multiplication and addition. Only here we work with boolean variables and AND (multiplication) and OR (addition) operations.
A + A' = 0 (when A = 0, A' = 0 and when A = 1, A' = 0 so (A + A') has to be 1)
2.a) Let's first derive the complement of F:
F' = (x'y + xyz')' = (x'y)'(xyz')' = (x + y')((xy)' + z) = (x + y')(x' + y' + z) = xx' + xy' + xz + x'y' + y'y' + y'z = 0 + xy' + xz + x'y' + y' + y'z = xy' + xz + y'(x + 1) + y'z = xy' + xz + y' + y'z = xy' + xz + y'(z + 1) = xy' + y' + xz = y'(x + 1) = xz + y'.
There is only one additional fact that I used here, that for any boolean variables A and B following holds:
A + AB = A(B + 1) = A - logically, variable A completely determines the output of such an expression and part AB cannot change the output of entire expression (you can check this one with truth tables if it's not clear, but boolean algebra should be enough to understand). And of course, for any boolean variable A + 1 = A.
2.b) FF' = (x'y + xyz')(xz + y') = x'yxz + x'yy' + xyz'xz + xyz'y'.
x'yxz = (xx')yz = 0xz = 0
xyy'= x0 = 0
xyz'xz = xxy(zz') = xy0 = 0
xyz'y' = xz'(yy') = xz'0 = 0
Therefore, FF' = 0.
2.c) F + F' = x'y + xyz' + xz + y'
This one is not so obvious. Let's start with two middle components and see what we can work out:
xyz' + xz = x(yz' + z) = x(yz' + z(y + y')) = x(yz' + yz + y'z) = x(y(z + z') + y'z) = x(y + y'z) = xy + xy'z.
I used the fact that we can write any boolean variable A in the following way:
A = A(B + B') = AB + AB' as (B + B') evaluates to 1 for any boolean variable B, so initial expression is not changed by AND-ing it together with such an expression.
Plugging this back in F + F' expression yields:
x'y + xy + xy'z + y' = y(x + x') + y'(xz + 1) = y + y' = 1.
I think I missed reading a theorem or postulate or something..
The equation is: wy + wxz + xyz
According to my professor, the simplification is (which she didn't explain how):
wy + xz(w'y + wy')
= wy + xz (w XOR y)
Where did that (w'y + wy') came from??
I tried to calculate it but so far I only got: (w+x)(w+y)(w+z)(x+y)(y+z)
In a Boolean expression + is XOR and * is AND. This is the same as identifying true with 1 and false with 0, with the only special convention that 1 + 1 = 0 (or 2 = 0 if you wish.)
With these definitions both + and * are commutative, i.e., a + b = b + a and a * b = b * a. In addition, the distributive law is also valid a * (b + c) = a * b + a * c. Note also that the * operator is usually implicit in the sense that we write ab instead of a * b.
Applying these properties to the expression wy + wxz + xyz, there are some few obvious transformations you can do:
wy + wxz + yxz (commute x with y)
wy + (w + y)xz (prev plus distribute xz)
wy + xz(w + y) (prev plus commute (w + y) with xz
Note that the last one is wy + xz(w XOR y) because + is nothing but XOR.
ADDENDUM
Regarding the expression of your professor, let's recall that a' = 1 - a by definition. So
w'y + wy' = (1 - w)y + w(1 - y) - def
= y - wy + w - wy - distribute
= y + w - simplify (a + a = 0 always)
= w + y - commute
which shows that s/he was right.
I not understand why the first boolean expression on the question can be simplified into the last. Please help me.
XY + Z(X ⊕ Y)
= XY + Z(X¬Y + ¬XY) // Expand XOR operation
= XY + X¬YZ + ¬XYZ // Distribute AND over OR
= XYZ + XY¬Z + X¬YZ + ¬XYZ // Expand XY
= (XYZ + XY¬Z) + (XYZ + X¬YZ) + (XYZ + ¬XYZ) // Copy XYZ and add parens
= XY + XZ + YZ // Remove trivial X+¬X = 1's
Reassembly is the reverse of disassembly.