How Does $AB{sim} C + {sim}ABC + {sim} A {sim}B{sim} C$ Turn into $(A+C+{sim}B)(B+{sim}A)(B+{sim} C)({sim}...

I'm trying to figure out this Boolean algebra question and I cannot for the life of me figure it out. I know that the answer is $(A+C+{sim} B)(B+{sim}A)(B+{sim} C)({sim} A+ {sim} C)$ but I can't find the steps to take it there from the original question of AB∼C+∼ABC+∼A∼B∼C.

Figured I would see if any of you knew.

Thanks

edited Jan 19 at 3:19

asked Jan 18 at 23:18

Adam

$begingroup$
There are only eight possible values for (A, B, C). Check all of them. I don't really understand what this question is asking for; if the two expressions are equal for all possible values, then they're equal for all possible values; that's the only thing you need to check to show equality.
$endgroup$
– Eric Lippert
Jan 18 at 23:21

1

$begingroup$
Is your question "what series of algebraic transformations transforms the first expression into the second?" If so, what techniques have you tried already?
$endgroup$
– Eric Lippert
Jan 18 at 23:27

$begingroup$
Yes, Eric. I have to prove that they are equal algebraically. I have tried factoring out different values among other things but that didn't seem to simplify it, I've used everything I can think of but haven't been able to crack it. any help would be greatly appreciated.
$endgroup$
– Adam
Jan 18 at 23:34

$begingroup$
Got it. Offhand I don't see it, but I'll play around with it for a bit.
$endgroup$
– Eric Lippert
Jan 19 at 0:19

add a comment |

Figured I would see if any of you knew.

Thanks

edited Jan 19 at 3:19

asked Jan 18 at 23:18

Adam

$begingroup$
There are only eight possible values for (A, B, C). Check all of them. I don't really understand what this question is asking for; if the two expressions are equal for all possible values, then they're equal for all possible values; that's the only thing you need to check to show equality.
$endgroup$
– Eric Lippert
Jan 18 at 23:21

1

$begingroup$
Is your question "what series of algebraic transformations transforms the first expression into the second?" If so, what techniques have you tried already?
$endgroup$
– Eric Lippert
Jan 18 at 23:27

$begingroup$
Yes, Eric. I have to prove that they are equal algebraically. I have tried factoring out different values among other things but that didn't seem to simplify it, I've used everything I can think of but haven't been able to crack it. any help would be greatly appreciated.
$endgroup$
– Adam
Jan 18 at 23:34

$begingroup$
Got it. Offhand I don't see it, but I'll play around with it for a bit.
$endgroup$
– Eric Lippert
Jan 19 at 0:19

add a comment |

Figured I would see if any of you knew.

Thanks

edited Jan 19 at 3:19

asked Jan 18 at 23:18

Adam

Figured I would see if any of you knew.

Thanks

logic propositional-calculus boolean-algebra

edited Jan 19 at 3:19

asked Jan 18 at 23:18

Adam

edited Jan 19 at 3:19

asked Jan 18 at 23:18

Adam

edited Jan 19 at 3:19

asked Jan 18 at 23:18

Adam

asked Jan 18 at 23:18

Adam

asked Jan 18 at 23:18

Adam

$begingroup$
There are only eight possible values for (A, B, C). Check all of them. I don't really understand what this question is asking for; if the two expressions are equal for all possible values, then they're equal for all possible values; that's the only thing you need to check to show equality.
$endgroup$
– Eric Lippert
Jan 18 at 23:21

1

$begingroup$
Is your question "what series of algebraic transformations transforms the first expression into the second?" If so, what techniques have you tried already?
$endgroup$
– Eric Lippert
Jan 18 at 23:27

$begingroup$
Yes, Eric. I have to prove that they are equal algebraically. I have tried factoring out different values among other things but that didn't seem to simplify it, I've used everything I can think of but haven't been able to crack it. any help would be greatly appreciated.
$endgroup$
– Adam
Jan 18 at 23:34

$begingroup$
Got it. Offhand I don't see it, but I'll play around with it for a bit.
$endgroup$
– Eric Lippert
Jan 19 at 0:19

add a comment |

$begingroup$
There are only eight possible values for (A, B, C). Check all of them. I don't really understand what this question is asking for; if the two expressions are equal for all possible values, then they're equal for all possible values; that's the only thing you need to check to show equality.
$endgroup$
– Eric Lippert
Jan 18 at 23:21

1

$begingroup$
Is your question "what series of algebraic transformations transforms the first expression into the second?" If so, what techniques have you tried already?
$endgroup$
– Eric Lippert
Jan 18 at 23:27

$begingroup$
Yes, Eric. I have to prove that they are equal algebraically. I have tried factoring out different values among other things but that didn't seem to simplify it, I've used everything I can think of but haven't been able to crack it. any help would be greatly appreciated.
$endgroup$
– Adam
Jan 18 at 23:34

$begingroup$
Got it. Offhand I don't see it, but I'll play around with it for a bit.
$endgroup$
– Eric Lippert
Jan 19 at 0:19

There are only eight possible values for (A, B, C). Check all of them. I don't really understand what this question is asking for; if the two expressions are equal for all possible values, then they're equal for all possible values; that's the only thing you need to check to show equality.

– Eric Lippert
Jan 18 at 23:21

Is your question "what series of algebraic transformations transforms the first expression into the second?" If so, what techniques have you tried already?

– Eric Lippert
Jan 18 at 23:27

Yes, Eric. I have to prove that they are equal algebraically. I have tried factoring out different values among other things but that didn't seem to simplify it, I've used everything I can think of but haven't been able to crack it. any help would be greatly appreciated.

– Adam
Jan 18 at 23:34

Got it. Offhand I don't see it, but I'll play around with it for a bit.

– Eric Lippert
Jan 19 at 0:19

add a comment |

2 Answers
2

active

oldest

votes

Attack the problem from the right side.

The distributive law is $(X + Y) Z = XZ + YZ$

Apply this law to $(A+C+{sim}B)(B+∼A)(B+{sim}C)(∼A+{sim}C)$ until you can no longer. You will end up with a very long expression of the form $ABB{sim}A + A{sim}AB{sim}A + ... + {sim}B{sim}A{sim}C{sim}C$

Then simplify that using the laws:

$X{sim}X = 0$

$X + X = X$

$XX = X$

$0 + X = X$

$0 X = 0$

$XY = YX$

$X + Y = Y + X$

and what will come out the other end is the left expression.

If you want to go from the left expression to the right expression, well, now you've got a set of steps that goes from right to left, so just reverse it.

edited Jan 19 at 0:54

answered Jan 19 at 0:49

Eric Lippert

3,2111418

add a comment |

Do you remember FOIL from algebra when dealing with numbers? It's that:

$$(a+b)(c+d)=ac+ad+bc+bd$$

Well, the same holds in a Boolean algebra. Moreover, it generalizes:

$$(a+b)(c+d+e)=ac+ad+ae+bc+bd+be$$

See how that works? You just multiply one variable from the one term with each one of the variables from the other term.

Notice how we can apply this to the first two terms of your RHS (right hand side):

$$(A + C +B')(B+A')=AB+AA'+CB+CA'+B'B+B'A'$$

(p.s. when working with $+$ and $cdot$, I much prefer working with $P'$ for negation rather than ~)

Now, as Eric indicated, at this point you can simplify:

$$AB+AA'+CB+CA'+B'B+B'A'$$

$$=AB+0+CB+CA'+0+
B'A'$$

$$=AB+CB+CA'+B'A'$$

Ok, now multiply this with $B+C'$, again canceling any term where you get a variable and its negation ... and then with $A'+C'$

Now, it also holds that:

$$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$

Again, see how that works? You multiply each one variable from each one of the terms with each other: you get each possible combination.

So, you could have done something like this in one big ass huge step as well, starting with your RHS: you would have gotten $3cdot 2cdot 2 cdot 2=24$ terms .... but all that would simplify to just the three terms from your LHS.

Now, you asked whether you couldn't start with the LHS and end up with the HS. Well, first of all, if you can figure out how to go from the RHS to the LHS, then you can just reverse that process to go from LHS to RHS, since they are all equivalences.

However, you can also do this. As it turns out, in Boolean algebra it also holds that:

$$ab+cd=(a+c)(a+d)(b+c)(b+d)$$

This is a little less intuitive, because this of course isn't true for numbers (which is why both Eric and I recommended going from RHS to LHS), but again, in a boolean algebra it does hold true, and so you could start with your LHS and do this:

$$ABC'+A'BC+A'B'C'=(A+A'+A')(A+A'+B')(A+A'+C')(A+B+A) ....$$

And so now you would have gotten $3 cdot 3 cdot 3 =27$ terms, but this time you have that $A + A'=1$, and so any term with a variable and its negation disappears ... and you should end up with just the $4$ terms of your RHS.

edited Jan 19 at 13:25

answered Jan 19 at 3:15

Bram28

63.2k44793

$begingroup$
Thanks for the response Bram I really appreciate it. I need to go the opposite way though. (A+C+∼B)(B+∼A)(B+∼C)(∼A+∼C) is the answer. I need to find how AB∼C+∼ABC+∼A∼B∼C Equals it. I may have phrased the question badly.
$endgroup$
– Adam
Jan 19 at 3:18

1

$begingroup$
@Adam: This is the same technique I suggested in my answer. It doesn't matter what direction you start from. If you're trying to show that X = Y, you can manipulate Y to equal X, or X to equal Y. All the operations are reversible; that's a fundamental property of equality. If you're required to show it in one direction, you can literally just write your equalities starting from the bottom of the page and moving up, instead of starting from the top and moving down!
$endgroup$
– Eric Lippert
Jan 19 at 4:28

$begingroup$
@Adam As Eric says, it doesn't matter which side you start: since every step is an equivalence, you can always reverse direction when you are all done. But, let me add something to my answer that shows you can start from the left as well.
$endgroup$
– Bram28
Jan 19 at 13:00

$begingroup$
@EricLippert Thanks Eric!
$endgroup$
– Bram28
Jan 19 at 13:25

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3078857%2fhow-does-ab-sim-c-simabc-sim-a-simb-sim-c-turn-into-ac-s%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Attack the problem from the right side.

The distributive law is $(X + Y) Z = XZ + YZ$

Then simplify that using the laws:

$X{sim}X = 0$

$X + X = X$

$XX = X$

$0 + X = X$

$0 X = 0$

$XY = YX$

$X + Y = Y + X$

and what will come out the other end is the left expression.

If you want to go from the left expression to the right expression, well, now you've got a set of steps that goes from right to left, so just reverse it.

edited Jan 19 at 0:54

answered Jan 19 at 0:49

Eric Lippert

3,2111418

add a comment |

Attack the problem from the right side.

The distributive law is $(X + Y) Z = XZ + YZ$

Then simplify that using the laws:

$X{sim}X = 0$

$X + X = X$

$XX = X$

$0 + X = X$

$0 X = 0$

$XY = YX$

$X + Y = Y + X$

and what will come out the other end is the left expression.

If you want to go from the left expression to the right expression, well, now you've got a set of steps that goes from right to left, so just reverse it.

edited Jan 19 at 0:54

answered Jan 19 at 0:49

Eric Lippert

3,2111418

add a comment |

Attack the problem from the right side.

The distributive law is $(X + Y) Z = XZ + YZ$

Then simplify that using the laws:

$X{sim}X = 0$

$X + X = X$

$XX = X$

$0 + X = X$

$0 X = 0$

$XY = YX$

$X + Y = Y + X$

and what will come out the other end is the left expression.

If you want to go from the left expression to the right expression, well, now you've got a set of steps that goes from right to left, so just reverse it.

edited Jan 19 at 0:54

answered Jan 19 at 0:49

Eric Lippert

3,2111418

Attack the problem from the right side.

The distributive law is $(X + Y) Z = XZ + YZ$

Then simplify that using the laws:

$X{sim}X = 0$

$X + X = X$

$XX = X$

$0 + X = X$

$0 X = 0$

$XY = YX$

$X + Y = Y + X$

and what will come out the other end is the left expression.

If you want to go from the left expression to the right expression, well, now you've got a set of steps that goes from right to left, so just reverse it.

edited Jan 19 at 0:54

answered Jan 19 at 0:49

Eric Lippert

3,2111418

edited Jan 19 at 0:54

answered Jan 19 at 0:49

Eric Lippert

3,2111418

answered Jan 19 at 0:49

Eric Lippert

3,2111418

answered Jan 19 at 0:49

Eric Lippert

3,2111418

add a comment |

Do you remember FOIL from algebra when dealing with numbers? It's that:

$$(a+b)(c+d)=ac+ad+bc+bd$$

Well, the same holds in a Boolean algebra. Moreover, it generalizes:

$$(a+b)(c+d+e)=ac+ad+ae+bc+bd+be$$

See how that works? You just multiply one variable from the one term with each one of the variables from the other term.

Notice how we can apply this to the first two terms of your RHS (right hand side):

$$(A + C +B')(B+A')=AB+AA'+CB+CA'+B'B+B'A'$$

(p.s. when working with $+$ and $cdot$, I much prefer working with $P'$ for negation rather than ~)

Now, as Eric indicated, at this point you can simplify:

$$AB+AA'+CB+CA'+B'B+B'A'$$

$$=AB+0+CB+CA'+0+
B'A'$$

$$=AB+CB+CA'+B'A'$$

Ok, now multiply this with $B+C'$, again canceling any term where you get a variable and its negation ... and then with $A'+C'$

Now, it also holds that:

$$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$

Again, see how that works? You multiply each one variable from each one of the terms with each other: you get each possible combination.

However, you can also do this. As it turns out, in Boolean algebra it also holds that:

$$ab+cd=(a+c)(a+d)(b+c)(b+d)$$

$$ABC'+A'BC+A'B'C'=(A+A'+A')(A+A'+B')(A+A'+C')(A+B+A) ....$$

edited Jan 19 at 13:25

answered Jan 19 at 3:15

Bram28

63.2k44793

$begingroup$
Thanks for the response Bram I really appreciate it. I need to go the opposite way though. (A+C+∼B)(B+∼A)(B+∼C)(∼A+∼C) is the answer. I need to find how AB∼C+∼ABC+∼A∼B∼C Equals it. I may have phrased the question badly.
$endgroup$
– Adam
Jan 19 at 3:18

1

$begingroup$
@Adam: This is the same technique I suggested in my answer. It doesn't matter what direction you start from. If you're trying to show that X = Y, you can manipulate Y to equal X, or X to equal Y. All the operations are reversible; that's a fundamental property of equality. If you're required to show it in one direction, you can literally just write your equalities starting from the bottom of the page and moving up, instead of starting from the top and moving down!
$endgroup$
– Eric Lippert
Jan 19 at 4:28

$begingroup$
@Adam As Eric says, it doesn't matter which side you start: since every step is an equivalence, you can always reverse direction when you are all done. But, let me add something to my answer that shows you can start from the left as well.
$endgroup$
– Bram28
Jan 19 at 13:00

$begingroup$
@EricLippert Thanks Eric!
$endgroup$
– Bram28
Jan 19 at 13:25

add a comment |

Do you remember FOIL from algebra when dealing with numbers? It's that:

$$(a+b)(c+d)=ac+ad+bc+bd$$

Well, the same holds in a Boolean algebra. Moreover, it generalizes:

$$(a+b)(c+d+e)=ac+ad+ae+bc+bd+be$$

See how that works? You just multiply one variable from the one term with each one of the variables from the other term.

Notice how we can apply this to the first two terms of your RHS (right hand side):

$$(A + C +B')(B+A')=AB+AA'+CB+CA'+B'B+B'A'$$

(p.s. when working with $+$ and $cdot$, I much prefer working with $P'$ for negation rather than ~)

Now, as Eric indicated, at this point you can simplify:

$$AB+AA'+CB+CA'+B'B+B'A'$$

$$=AB+0+CB+CA'+0+
B'A'$$

$$=AB+CB+CA'+B'A'$$

Ok, now multiply this with $B+C'$, again canceling any term where you get a variable and its negation ... and then with $A'+C'$

Now, it also holds that:

$$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$

Again, see how that works? You multiply each one variable from each one of the terms with each other: you get each possible combination.

However, you can also do this. As it turns out, in Boolean algebra it also holds that:

$$ab+cd=(a+c)(a+d)(b+c)(b+d)$$

$$ABC'+A'BC+A'B'C'=(A+A'+A')(A+A'+B')(A+A'+C')(A+B+A) ....$$

edited Jan 19 at 13:25

answered Jan 19 at 3:15

Bram28

63.2k44793

$begingroup$
Thanks for the response Bram I really appreciate it. I need to go the opposite way though. (A+C+∼B)(B+∼A)(B+∼C)(∼A+∼C) is the answer. I need to find how AB∼C+∼ABC+∼A∼B∼C Equals it. I may have phrased the question badly.
$endgroup$
– Adam
Jan 19 at 3:18

1

$begingroup$
@Adam: This is the same technique I suggested in my answer. It doesn't matter what direction you start from. If you're trying to show that X = Y, you can manipulate Y to equal X, or X to equal Y. All the operations are reversible; that's a fundamental property of equality. If you're required to show it in one direction, you can literally just write your equalities starting from the bottom of the page and moving up, instead of starting from the top and moving down!
$endgroup$
– Eric Lippert
Jan 19 at 4:28

$begingroup$
@Adam As Eric says, it doesn't matter which side you start: since every step is an equivalence, you can always reverse direction when you are all done. But, let me add something to my answer that shows you can start from the left as well.
$endgroup$
– Bram28
Jan 19 at 13:00

$begingroup$
@EricLippert Thanks Eric!
$endgroup$
– Bram28
Jan 19 at 13:25

add a comment |

Do you remember FOIL from algebra when dealing with numbers? It's that:

$$(a+b)(c+d)=ac+ad+bc+bd$$

Well, the same holds in a Boolean algebra. Moreover, it generalizes:

$$(a+b)(c+d+e)=ac+ad+ae+bc+bd+be$$

See how that works? You just multiply one variable from the one term with each one of the variables from the other term.

Notice how we can apply this to the first two terms of your RHS (right hand side):

$$(A + C +B')(B+A')=AB+AA'+CB+CA'+B'B+B'A'$$

(p.s. when working with $+$ and $cdot$, I much prefer working with $P'$ for negation rather than ~)

Now, as Eric indicated, at this point you can simplify:

$$AB+AA'+CB+CA'+B'B+B'A'$$

$$=AB+0+CB+CA'+0+
B'A'$$

$$=AB+CB+CA'+B'A'$$

Ok, now multiply this with $B+C'$, again canceling any term where you get a variable and its negation ... and then with $A'+C'$

Now, it also holds that:

$$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$

Again, see how that works? You multiply each one variable from each one of the terms with each other: you get each possible combination.

However, you can also do this. As it turns out, in Boolean algebra it also holds that:

$$ab+cd=(a+c)(a+d)(b+c)(b+d)$$

$$ABC'+A'BC+A'B'C'=(A+A'+A')(A+A'+B')(A+A'+C')(A+B+A) ....$$

edited Jan 19 at 13:25

answered Jan 19 at 3:15

Bram28

63.2k44793

Do you remember FOIL from algebra when dealing with numbers? It's that:

$$(a+b)(c+d)=ac+ad+bc+bd$$

Well, the same holds in a Boolean algebra. Moreover, it generalizes:

$$(a+b)(c+d+e)=ac+ad+ae+bc+bd+be$$

See how that works? You just multiply one variable from the one term with each one of the variables from the other term.

Notice how we can apply this to the first two terms of your RHS (right hand side):

$$(A + C +B')(B+A')=AB+AA'+CB+CA'+B'B+B'A'$$

(p.s. when working with $+$ and $cdot$, I much prefer working with $P'$ for negation rather than ~)

Now, as Eric indicated, at this point you can simplify:

$$AB+AA'+CB+CA'+B'B+B'A'$$

$$=AB+0+CB+CA'+0+
B'A'$$

$$=AB+CB+CA'+B'A'$$

Ok, now multiply this with $B+C'$, again canceling any term where you get a variable and its negation ... and then with $A'+C'$

Now, it also holds that:

$$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$

Again, see how that works? You multiply each one variable from each one of the terms with each other: you get each possible combination.

However, you can also do this. As it turns out, in Boolean algebra it also holds that:

$$ab+cd=(a+c)(a+d)(b+c)(b+d)$$

$$ABC'+A'BC+A'B'C'=(A+A'+A')(A+A'+B')(A+A'+C')(A+B+A) ....$$

edited Jan 19 at 13:25

answered Jan 19 at 3:15

Bram28

63.2k44793

edited Jan 19 at 13:25

answered Jan 19 at 3:15

Bram28

63.2k44793

answered Jan 19 at 3:15

Bram28

63.2k44793

answered Jan 19 at 3:15

Bram28

63.2k44793

$begingroup$
Thanks for the response Bram I really appreciate it. I need to go the opposite way though. (A+C+∼B)(B+∼A)(B+∼C)(∼A+∼C) is the answer. I need to find how AB∼C+∼ABC+∼A∼B∼C Equals it. I may have phrased the question badly.
$endgroup$
– Adam
Jan 19 at 3:18

1

$begingroup$
@Adam: This is the same technique I suggested in my answer. It doesn't matter what direction you start from. If you're trying to show that X = Y, you can manipulate Y to equal X, or X to equal Y. All the operations are reversible; that's a fundamental property of equality. If you're required to show it in one direction, you can literally just write your equalities starting from the bottom of the page and moving up, instead of starting from the top and moving down!
$endgroup$
– Eric Lippert
Jan 19 at 4:28

$begingroup$
@Adam As Eric says, it doesn't matter which side you start: since every step is an equivalence, you can always reverse direction when you are all done. But, let me add something to my answer that shows you can start from the left as well.
$endgroup$
– Bram28
Jan 19 at 13:00

$begingroup$
@EricLippert Thanks Eric!
$endgroup$
– Bram28
Jan 19 at 13:25

add a comment |

$begingroup$
Thanks for the response Bram I really appreciate it. I need to go the opposite way though. (A+C+∼B)(B+∼A)(B+∼C)(∼A+∼C) is the answer. I need to find how AB∼C+∼ABC+∼A∼B∼C Equals it. I may have phrased the question badly.
$endgroup$
– Adam
Jan 19 at 3:18

1

$begingroup$
@Adam: This is the same technique I suggested in my answer. It doesn't matter what direction you start from. If you're trying to show that X = Y, you can manipulate Y to equal X, or X to equal Y. All the operations are reversible; that's a fundamental property of equality. If you're required to show it in one direction, you can literally just write your equalities starting from the bottom of the page and moving up, instead of starting from the top and moving down!
$endgroup$
– Eric Lippert
Jan 19 at 4:28

$begingroup$
@Adam As Eric says, it doesn't matter which side you start: since every step is an equivalence, you can always reverse direction when you are all done. But, let me add something to my answer that shows you can start from the left as well.
$endgroup$
– Bram28
Jan 19 at 13:00

$begingroup$
@EricLippert Thanks Eric!
$endgroup$
– Bram28
Jan 19 at 13:25

Thanks for the response Bram I really appreciate it. I need to go the opposite way though. (A+C+∼B)(B+∼A)(B+∼C)(∼A+∼C) is the answer. I need to find how AB∼C+∼ABC+∼A∼B∼C Equals it. I may have phrased the question badly.

– Adam
Jan 19 at 3:18

@Adam: This is the same technique I suggested in my answer. It doesn't matter what direction you start from. If you're trying to show that X = Y, you can manipulate Y to equal X, or X to equal Y. All the operations are reversible; that's a fundamental property of equality. If you're required to show it in one direction, you can literally just write your equalities starting from the bottom of the page and moving up, instead of starting from the top and moving down!

– Eric Lippert
Jan 19 at 4:28

@Adam As Eric says, it doesn't matter which side you start: since every step is an equivalence, you can always reverse direction when you are all done. But, let me add something to my answer that shows you can start from the left as well.

– Bram28
Jan 19 at 13:00

@EricLippert Thanks Eric!

– Bram28
Jan 19 at 13:25

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Mathematics Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu