Which laws do I need to use to simplify
!X + (!Y + !Z)*(Y + Z)
to
!X + (Y*!Z) + (!Y*Z)
No need to apply any specific 'laws', just write down a simple table to figure out the two expression are equivalent (providing + is OR and * is AND):
Y Z !Y !Z (!Y+!Z)*(Y+Z) (Y*!Z) +(!Y*Z)
0 0 1 1 (1+1=1)*(0+0=0)=0 (0*1=0)+(1*0=0)=0
0 1 1 0 (1+0=1)*(0+1=1)=1 (0*0=0)+(1*1=1)=1
1 0 0 1 (0+1=1)*(1+0=1)=1 (1*1=1)+(0*0=0)=1
1 1 0 0 (0+0=0)*(1+1=1)=0 (1*0=0)+(0*1=0)=0
Likewise you can deduct that (X or Y) and Z is equivalent to (X and Z) or (Y and Z). This is called distributivity of and over or. Further reading is a nice article about Boolean algebra on Wikipedia.
In your example: (!Y + !Z)*(Y + Z) = !Y*(Y + Z) + !Z*(Y + Z) = !Y*Y + !Y*Z + !Z*Y + !Z*Z. Trivially A and not A == false, then your expression simplifies to !Y*Z + !Z*Y and further to Y*!Z + !Y*Z because of commutativity.
You can first start by distributivity of multiplication over addition:
!X + (!Y + !Z)*(Y + Z) = !X + !Y*Y + !Z*Y + !Y*Z + !Z*Z
Then, we can use complementation to remove elements of the form !p*p:
= !X + 0 + !Z*Y + !Y*Z + 0
And finally remove the 0 as they are +'s neutral.
If memory serves (and I take your notation correctly), then that's an SLD resolution inside a simple conjunction: http://en.wikipedia.org/wiki/SLD_resolution
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'm having trouble simplifying these two boolean expressions algebraically and proving them with a Karnaugh Map. How can I do this?
These are my two expressions:
1) (X * Y) + (X' * Y * Z')+ (Y * Z)
2) (X * Y') + Z + (X' + Y)+ (Y * Z)
I've tried going through it using the boolean theorems and laws to reduce them but I always come up with different answers. My answers usually comes up as this.
1) (Y * Z') + (X' * Y)
2) (X' * Y') + (X * Y' * Z')
I don't know if my K-Map is wrong, but I do need someone to help me understand how to solve this problem and the steps or laws I need to get the answer, so that I can master it. It is practice for exam, and I suck at boolean algebra. I appreciate it.
Let's start with the first expression:
E = XY + X'YZ' + YZ
The three terms have Y, then we can factor it out
E = Y(X + X'Z' + Z)
Now let's concentrate in the expression in parenthesis S = X + X'Z' + Z:
S = X + X'Z' + Z
= X + (X + Z)' + Z (De Morgan)
= (X + Z) + (X + Z)' (regrouping)
so, despite the fact that this still looks complex it has the form
S = p + p'
for p = X + Z, right? But p + p' = 1 (or true) no matter the value of p. Thus the expression S is 1 and we get
E = Y(X + X'Z' + Z) = YS = Y1 = Y
In other words, the first expression reduces to Y.
Notice also that it is not that hard to see why S = 1 without rewriting it. There are three cases: (a) If X is true, then certainly the expression is true. (b) If Z is true, the result is true also. (c) If none of X and Z are true then both are false and X'Z' is true. So, in each of these 3 cases at least one of the terms is true, hence their sum.
Let's now consider the second expression
F = XY' + Z + (X' + Y) + YZ
The first thing to note is that XY' is the opposite of (X' + Y):
(X' + Y) = (XY')' (De Morgan)
So,
F = XY' + (XY')' + Z + YZ
Again, regardless of the fact that XY' + (XY')' looks complicated, it is an expression of the form p + p'. But p + p' = 1 (it is always true) and therefore
F = 1 + Z + YZ = 1
no matter the values of Y and Z. So, the second expression is nothing but 1 (aka true).
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.
Hi I'v solved this halfway, please help me for the rest.
so far I've got..
x'yz + xy'z + xyz' + xyz
z(x'y + xy') + xy(z'+z)
z(x'y + xy') + xy
i don't understand how to solve the z(x'y + xy') part of this expression.. please somebody help..
x'y + xy' is XOR. So you can simplify to z(x+y) + xy because z(x'y + xy') + xy is z when x != y and xy when x == y.
The (x+y) after z is necessary to disallow influence of z when x == y == 0.