Mixing
How do I get Z' + X'Y' + XY from Z' + ZX'Y' + ZXY?
$begingroup$
So I'm trying to simplify an equation using boolean algebra. I'm still very confused as to how all the rules work and such, but I feel like I'm very close to the answer.
The original equation is (in canonical SOP form):
Z'X'Y' + Z'X'Y + Z'XY' + Z'XY + ZX'Y' + ZXY
So far I've gotten this far:
Distributive Rule 12a: Z'X'(Y + Y') + Z'X(Y' + Y) + ZX'Y' + ZXY
Identity Rule 6a: Z'X' + Z'X + ZX'Y' + ZXY
Rule 12a again: Z'(X' + X) + ZX'Y' + ZXY
Rule 6a again: Z' + ZX'Y' + ZXY
My issue is that the answer is Z' + X'Y' + XY, but how do I get there? Because doesn't ZX'Y' + ZXY simplify to Z(X'Y' + XY) = Z?
What am I doing wrong? Are the steps I have so far correct?
Thanks.
boolean-algebra
asked Jan 31 at 5:13
ESMESM
104
$endgroup$
add a comment |
$begingroup$
So I'm trying to simplify an equation using boolean algebra. I'm still very confused as to how all the rules work and such, but I feel like I'm very close to the answer.
The original equation is (in canonical SOP form):
Z'X'Y' + Z'X'Y + Z'XY' + Z'XY + ZX'Y' + ZXY
So far I've gotten this far:
Distributive Rule 12a: Z'X'(Y + Y') + Z'X(Y' + Y) + ZX'Y' + ZXY
Identity Rule 6a: Z'X' + Z'X + ZX'Y' + ZXY
Rule 12a again: Z'(X' + X) + ZX'Y' + ZXY
Rule 6a again: Z' + ZX'Y' + ZXY
My issue is that the answer is Z' + X'Y' + XY, but how do I get there? Because doesn't ZX'Y' + ZXY simplify to Z(X'Y' + XY) = Z?
What am I doing wrong? Are the steps I have so far correct?
Thanks.
boolean-algebra
asked Jan 31 at 5:13
ESMESM
104
$endgroup$
$begingroup$
Isn't it supposed to be in capitals in the title?
$endgroup$
– Mohammad Zuhair Khan
Jan 31 at 5:16
$begingroup$
@MohammadZuhairKhan fixed.
$endgroup$
– ESM
Jan 31 at 5:20
add a comment |
$begingroup$
So I'm trying to simplify an equation using boolean algebra. I'm still very confused as to how all the rules work and such, but I feel like I'm very close to the answer.
The original equation is (in canonical SOP form):
Z'X'Y' + Z'X'Y + Z'XY' + Z'XY + ZX'Y' + ZXY
So far I've gotten this far:
Distributive Rule 12a: Z'X'(Y + Y') + Z'X(Y' + Y) + ZX'Y' + ZXY
Identity Rule 6a: Z'X' + Z'X + ZX'Y' + ZXY
Rule 12a again: Z'(X' + X) + ZX'Y' + ZXY
Rule 6a again: Z' + ZX'Y' + ZXY
My issue is that the answer is Z' + X'Y' + XY, but how do I get there? Because doesn't ZX'Y' + ZXY simplify to Z(X'Y' + XY) = Z?
What am I doing wrong? Are the steps I have so far correct?
Thanks.
boolean-algebra
asked Jan 31 at 5:13
ESMESM
104
$endgroup$
So I'm trying to simplify an equation using boolean algebra. I'm still very confused as to how all the rules work and such, but I feel like I'm very close to the answer.
The original equation is (in canonical SOP form):
Z'X'Y' + Z'X'Y + Z'XY' + Z'XY + ZX'Y' + ZXY
So far I've gotten this far:
Distributive Rule 12a: Z'X'(Y + Y') + Z'X(Y' + Y) + ZX'Y' + ZXY
Identity Rule 6a: Z'X' + Z'X + ZX'Y' + ZXY
Rule 12a again: Z'(X' + X) + ZX'Y' + ZXY
Rule 6a again: Z' + ZX'Y' + ZXY
My issue is that the answer is Z' + X'Y' + XY, but how do I get there? Because doesn't ZX'Y' + ZXY simplify to Z(X'Y' + XY) = Z?
What am I doing wrong? Are the steps I have so far correct?
Thanks.
boolean-algebra
boolean-algebra
asked Jan 31 at 5:13
ESMESM
104
asked Jan 31 at 5:13
ESMESM
104
edited Jan 31 at 5:20
ESM
asked Jan 31 at 5:13
ESMESM
104
asked Jan 31 at 5:13
ESMESM
104
asked Jan 31 at 5:13
ESMESM
104
104
$begingroup$
Isn't it supposed to be in capitals in the title?
$endgroup$
– Mohammad Zuhair Khan
Jan 31 at 5:16
$begingroup$
@MohammadZuhairKhan fixed.
$endgroup$
– ESM
Jan 31 at 5:20
add a comment |
$begingroup$
Isn't it supposed to be in capitals in the title?
$endgroup$
– Mohammad Zuhair Khan
Jan 31 at 5:16
$begingroup$
@MohammadZuhairKhan fixed.
$endgroup$
– ESM
Jan 31 at 5:20
$begingroup$
Isn't it supposed to be in capitals in the title?
$endgroup$
– Mohammad Zuhair Khan
Jan 31 at 5:16
$begingroup$
Isn't it supposed to be in capitals in the title?
$endgroup$
– Mohammad Zuhair Khan
Jan 31 at 5:16
$begingroup$
@MohammadZuhairKhan fixed.
$endgroup$
– ESM
Jan 31 at 5:20
$begingroup$
@MohammadZuhairKhan fixed.
$endgroup$
– ESM
Jan 31 at 5:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because doesn't $ZX'Y'+ZXY$ simplify to $Z(X'Y'+XY)=Z$?
No, because $(XY)'ne X'Y'$, so in parentheses you do NOT have something of the form "$A+A'$". (In fact, by De Morgan's laws, $(XY)'=X'+Y'$.)
My issue is that the answer is $Z'+X'Y'+XY$, but how do I get there?
There are some additional useful properties that often help simplify boolean expressions. Here are two of them.
1) $color{red}{A+AB=A}$. Proof:
$$A+AB=A1+AB=A(1+B)=A1=A.$$
2) $color{red}{A+A'B=A+B}$. Proof:
$$A+A'B=A+AB+A'B=A+(A+A')B=A+1B=A+B.$$
Now, $Z'+ZX'Y'+ZXY=Z'+Z(X'Y'+XY)=Z'+X'Y'+XY$, where the last step is an application of the second property above (with $A=Z'$ and $B=X'Y'+XY$).
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
$endgroup$
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
add a comment |
$begingroup$
You need to immediately add the following equivalence to your toolbox:
Reduction
$P + P'Q=P+Q$
That is, the $P$ term 'reduces' the $P'Q$ term to just $Q$
With that, you can see how in your expression, the $Z'$ term reduces the $ZX'Y'$ to $X'Y'$, and also reduces the $ZXY$ to $XY$. So there:
$Z'+ZX'Y'+ZXY overset{Reduction x 2}{=} Z'+X'Y'+XY$
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Because doesn't $ZX'Y'+ZXY$ simplify to $Z(X'Y'+XY)=Z$?
No, because $(XY)'ne X'Y'$, so in parentheses you do NOT have something of the form "$A+A'$". (In fact, by De Morgan's laws, $(XY)'=X'+Y'$.)
My issue is that the answer is $Z'+X'Y'+XY$, but how do I get there?
There are some additional useful properties that often help simplify boolean expressions. Here are two of them.
1) $color{red}{A+AB=A}$. Proof:
$$A+AB=A1+AB=A(1+B)=A1=A.$$
2) $color{red}{A+A'B=A+B}$. Proof:
$$A+A'B=A+AB+A'B=A+(A+A')B=A+1B=A+B.$$
Now, $Z'+ZX'Y'+ZXY=Z'+Z(X'Y'+XY)=Z'+X'Y'+XY$, where the last step is an application of the second property above (with $A=Z'$ and $B=X'Y'+XY$).
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
$endgroup$
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
add a comment |
$begingroup$
Because doesn't $ZX'Y'+ZXY$ simplify to $Z(X'Y'+XY)=Z$?
No, because $(XY)'ne X'Y'$, so in parentheses you do NOT have something of the form "$A+A'$". (In fact, by De Morgan's laws, $(XY)'=X'+Y'$.)
My issue is that the answer is $Z'+X'Y'+XY$, but how do I get there?
There are some additional useful properties that often help simplify boolean expressions. Here are two of them.
1) $color{red}{A+AB=A}$. Proof:
$$A+AB=A1+AB=A(1+B)=A1=A.$$
2) $color{red}{A+A'B=A+B}$. Proof:
$$A+A'B=A+AB+A'B=A+(A+A')B=A+1B=A+B.$$
Now, $Z'+ZX'Y'+ZXY=Z'+Z(X'Y'+XY)=Z'+X'Y'+XY$, where the last step is an application of the second property above (with $A=Z'$ and $B=X'Y'+XY$).
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
$endgroup$
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
add a comment |
$begingroup$
Because doesn't $ZX'Y'+ZXY$ simplify to $Z(X'Y'+XY)=Z$?
No, because $(XY)'ne X'Y'$, so in parentheses you do NOT have something of the form "$A+A'$". (In fact, by De Morgan's laws, $(XY)'=X'+Y'$.)
My issue is that the answer is $Z'+X'Y'+XY$, but how do I get there?
There are some additional useful properties that often help simplify boolean expressions. Here are two of them.
1) $color{red}{A+AB=A}$. Proof:
$$A+AB=A1+AB=A(1+B)=A1=A.$$
2) $color{red}{A+A'B=A+B}$. Proof:
$$A+A'B=A+AB+A'B=A+(A+A')B=A+1B=A+B.$$
Now, $Z'+ZX'Y'+ZXY=Z'+Z(X'Y'+XY)=Z'+X'Y'+XY$, where the last step is an application of the second property above (with $A=Z'$ and $B=X'Y'+XY$).
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
$endgroup$
Because doesn't $ZX'Y'+ZXY$ simplify to $Z(X'Y'+XY)=Z$?
No, because $(XY)'ne X'Y'$, so in parentheses you do NOT have something of the form "$A+A'$". (In fact, by De Morgan's laws, $(XY)'=X'+Y'$.)
My issue is that the answer is $Z'+X'Y'+XY$, but how do I get there?
There are some additional useful properties that often help simplify boolean expressions. Here are two of them.
1) $color{red}{A+AB=A}$. Proof:
$$A+AB=A1+AB=A(1+B)=A1=A.$$
2) $color{red}{A+A'B=A+B}$. Proof:
$$A+A'B=A+AB+A'B=A+(A+A')B=A+1B=A+B.$$
Now, $Z'+ZX'Y'+ZXY=Z'+Z(X'Y'+XY)=Z'+X'Y'+XY$, where the last step is an application of the second property above (with $A=Z'$ and $B=X'Y'+XY$).
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
edited Jan 31 at 5:49
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
answered Jan 31 at 5:44
zipirovichzipirovich
11.3k11731
11.3k11731
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
add a comment |
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
$begingroup$
Ohhh I gotcha. Yea that must be using DeMorgan's Theorem then. I never thought to have the Z' and Z = A and the XY = B. Thank you so much man! That definitely helps to clear things up for me :)
$endgroup$
– ESM
Jan 31 at 5:55
add a comment |
$begingroup$
You need to immediately add the following equivalence to your toolbox:
Reduction
$P + P'Q=P+Q$
That is, the $P$ term 'reduces' the $P'Q$ term to just $Q$
With that, you can see how in your expression, the $Z'$ term reduces the $ZX'Y'$ to $X'Y'$, and also reduces the $ZXY$ to $XY$. So there:
$Z'+ZX'Y'+ZXY overset{Reduction x 2}{=} Z'+X'Y'+XY$
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
You need to immediately add the following equivalence to your toolbox:
Reduction
$P + P'Q=P+Q$
That is, the $P$ term 'reduces' the $P'Q$ term to just $Q$
With that, you can see how in your expression, the $Z'$ term reduces the $ZX'Y'$ to $X'Y'$, and also reduces the $ZXY$ to $XY$. So there:
$Z'+ZX'Y'+ZXY overset{Reduction x 2}{=} Z'+X'Y'+XY$
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
You need to immediately add the following equivalence to your toolbox:
Reduction
$P + P'Q=P+Q$
That is, the $P$ term 'reduces' the $P'Q$ term to just $Q$
With that, you can see how in your expression, the $Z'$ term reduces the $ZX'Y'$ to $X'Y'$, and also reduces the $ZXY$ to $XY$. So there:
$Z'+ZX'Y'+ZXY overset{Reduction x 2}{=} Z'+X'Y'+XY$
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
$endgroup$
You need to immediately add the following equivalence to your toolbox:
Reduction
$P + P'Q=P+Q$
That is, the $P$ term 'reduces' the $P'Q$ term to just $Q$
With that, you can see how in your expression, the $Z'$ term reduces the $ZX'Y'$ to $X'Y'$, and also reduces the $ZXY$ to $XY$. So there:
$Z'+ZX'Y'+ZXY overset{Reduction x 2}{=} Z'+X'Y'+XY$
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
answered Feb 1 at 3:24
Bram28Bram28
64.2k44793
64.2k44793
add a comment |
add a comment |
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$begingroup$
Isn't it supposed to be in capitals in the title?
$endgroup$
– Mohammad Zuhair Khan
Jan 31 at 5:16
$begingroup$
@MohammadZuhairKhan fixed.
$endgroup$
– ESM
Jan 31 at 5:20