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I have this truth table, and I must find the right formula. (T = TRUE , V = FALSE)

Where should I start with this?

I first wrote out everything :

A&B&C&D v A&B&C&-D v -A&-B&C&-D etc...

But then I was told, that I can solve this even easier. Any suggetions?

EDIT: the last column A is the truth value of the correct formula.

enter image description here

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I would title the last column something besides $A$ as you have already used $A$. Say we call it $F$. You might make a column of the truth value of $F \iff C$ and find some inspiration. – Ross Millikan Sep 28 '11 at 22:44
Related:… – lhf Sep 28 '11 at 22:53
If you compare the desired output to $A \iff C$ you are getting close. – Ross Millikan Sep 28 '11 at 23:05
I got $(a\land c)\lor (a\land \neg b\land d)\lor (\neg a\land b\land \neg c)\lor (\neg a\land b\land \neg d)\lor (\neg a\land \neg c\land \neg d)$ just in case you want to control your result. – Listing Sep 28 '11 at 23:05
up vote 2 down vote accepted

Solution of Karnaugh map:

enter image description here

$(A\land C)\lor (A\land \neg B\land D)\lor (\neg A\land B\land \neg C)\lor (\neg A\land B\land \neg D)\lor (\neg A\land \neg C\land \neg D)$

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Why does 0110 not connect with 1110? I connected them in my map...can you explain? Is my solution wrong, or there are two possibilities? – Jaanus Sep 29 '11 at 8:43
What my solution was : A&C v A&-B&D v B&C&-D v -A&-B&-D v -A&B&-C – Jaanus Sep 29 '11 at 8:59
@Jaanus,Yes there are two possibilities,both correct...but you have to use only one of them in final expression – pedja Sep 29 '11 at 9:06
So my solution is also correct, by the looks of it? – Jaanus Sep 29 '11 at 10:06
@Jaanus,it should be $B\land C\land \neg D$ instead $B\land \neg C\land \neg D$ – pedja Sep 29 '11 at 10:33

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