Boolean Expression Represented as a Truth Table Example 1

Hi guys, I'm Jonathan Lambert from the Mathematical Development and Support Service at the National College of Arlen's and this short video in the video series I started posting on my channel is about Boolean

expression

s and he's going to look at how to take a Boolean

expression

and represent the

boolean

expression in a

truth

table

and the construction of the expression that I have here has one two three four main terms and it seems a little complicated, but what we are going to do is We are going to take this particular expression and step by step term by term from left to right and we are going to represent it in a tree

table

.

The first thing we need to look at from this particular expression is the inputs to this particular from this particular system and how many unique letters are there in the system where we have an a and we have B and C, so this is a tree input system and so What we will do is for our

truth

table, don't forget. The actual table has two parts: the left part where we list the inputs and the possible states of the inputs coming in at any time and the right part of this particular bar, so let me make this a double bar, okay? where we have the evaluation of the

boolean

expression stage by stage, we will simply make that double slash to differentiate it from any other particular column that we are going to have in this table, so these three entries are B and C. so we have a a, we have a B and we have a C and what we need to do is list all the possible states that a B and a C could be in in this particular system and while C could be in so off could be 0 could be 1 will be 0 could be 1 could be 0 could be 1 could be 0 could be one we continue that way, okay, really what we want to do is try to finish a full roll warms up at the end of this particular representation, you see, so the first entry more close to the double bar alternates 0 1 0 1 0 1 0 1 and continues in this particular way forever, okay, the next entry has two zeros followed by two one two zeros followed by two ones and continues in that way forever okay , the next entry has four zeros zero zero zero zero followed by four ones 1 1 1 1 and we will continue that way yes, but we know when to stop here stop when we have a complete list at the end of the representation, ok, these are all the states possible to try that the 3 input system could be at any time and the question we have is that when we pass through the states this boolean expression, what will be the result, so to do this in a target, we will take the first term, ok, which is an AND bar with a bar a or B and we will do the representation of that first of all.

So I go from left to right before I do this and I have to calculate a bar, so let's create a column called bar or the negation of a, the negation of a is simply when a is 0, which is there, it becomes 1 . so 0 goes to 1 0 goes to 1 0 goes to 1 and when it's used it goes to 0 then 1 goes to 0 0 0 0 and that gives us our column and that will give us a column, a bar, okay, so there this. the first time you called me, you just need to finish with this expression here, okay, but this expression is an A or it would be Bart and Bart crosses the operator, so before we can do the inversion or the negation, we have to figure out what or would be like this, then let's make a column called a or would be fine to our VP is just the simple order of two inputs and we know that it never gives the 0 when both inputs simultaneously are 0, so you can see here that there are 0 so we had a 0 here there are 0 we have a zero they are not both simultaneously 0 so we had a 1 not both simultaneously 0 we get a 1 and so on they are not both simultaneously 0 so we are we are going to want so we end up with a or B okay, now that we have calculated our B we can go ahead and we can calculate the inversion of what is a or B bar okay, then the negation of ARB while ARB and the values ​​are here, so the negation of them zero goes to 1 0 goes to 1 1 goes to 0 0 0 0 0 which gives us a bar column a or b well and now that we have the bar column a or b and we have in the column a bar we can now undertake or perform the end of both particular sets of values, so what we want to calculate next is a bar and is with bar a or b, okay, don't forget that an and only gives us 1 when both are simultaneously 1 so here we have two ones so we get a 1 here we have 2 ones we get a 1 and another case here is 0 so we will never have simultaneous ones so we end up with 0 0 0 0 0 0 ok, so this term here now this column represents the first term in our boolean expression.

To continue, we need to calculate an AND with a or C, but before we do this out for out out around, we have to do this internal. or in case we need to calculate a o because it's just an order of column a with column C and let's not forget once again for an or and or just get 0 and both are simultaneously 0 so we have two 0's here like this that we get to 0, we don't have two zeros here, so we get a 1, we get two 0s here, so we get a 0, we don't have two zeros here, so we get a 1 and we don't get two. zeros here or here or here or here so we end up with 1 1 1 1 and that gives us a o with our C ok and now what we have to do is now we can do the I D a AND with a or C to give us our second term and then what we have to do is a final, don't forget a and just give us heat in both muscle tensions one, but look at our two columns here, column a, column a or C, then column a. is the first column and column a or C is the last delight in the last column we did for Nance or assimilators no no no no no yes yes yes yes, so the last four values ​​are simultaneous as if it were one in all the others places where we are we're going to have zero okay let's start with the next term but what we need to do is end with a slash or C so before we finish the outer end we have to do this term here before we can do the investment or the bar. we have to do it right and we have already calculated a o C here is the column here so we can calculate its inversion a o C bar V which is a false a o C column and convert it and invert it so that zero goes from one zero one zero zero zero zero zero to give us a o with C okay, we're almost there, now we can calculate what B ends with this particular value here a o with C bar, B is simply column B and a o with C The bar is the column of bar C of the aorta, so it's going to give us B ANDed with a or with bar C, so again it's an and the outer ends, so we just get a Baltimore woman with a similar heat, so there's room in B and a are bar C. column, so there is nothing similar, there is one here or here there is in the third case, so here is the one of the first two and everything else is 0, so we never have the simultaneous ones until the end and since we have zeros here, we are done. up with 0 0 0 0 0 so now we have calculated the Tudor term okay we are almost there and the next thing we need to calculate is we need to calculate a little more complicated okay we look at this term on the left side at the end and it is a bar to cross an o but before we can make that bar along your we have to calculate a bar, we also need to calculate a bar C, yes we have already calculated a bar before so we have that particular input so what we need a bar C, we haven't calculated it anywhere, so we will need a bar AC, which is taking column C and inverting it so that zero goes to 1 1 goes to 0 1 0 1 0 1 0 to give us our bar column C and now we can create our bar or corded with our bar column C, okay, our bar A, sort our six bar column, it's just a bar column, which is the second, the first column that we did well or well. the first application we did together with our C bar from the last kilometer, so in o it only gives us 0 and they are both simultaneously zero, so they are not similar.

I see zero here, it doesn't matter here or here or here or oops or here. Well, they are there. so we get a zero here, they are not simultaneously zero there, but they are in the last case here, so we end up with something that looks like this and there are 1 2 3 4 values ​​1 2 3 4 fighting, so this gives us a bar over bar C, so we've done what's inside or inside the inverter and our negation, and now we can negate it, so it's a bar over bar C. Look with your knees, okay, so we skip the 1 to 0, so this becomes 0 0 0 0. 0 1 0 1 bright.

I'll do it now, so we've calculated the left side of this particular and here okay, so we have to calculate what is a or B, but we've already calculated our a, our B. it's in the column okay, yesterday we can calculate how to use the rule here to cross this and that's a little bit clearer, in particular, what we can do is calculate what is a bar or bar C, bar which is this column that you need. to be handled with column A or B, which is this column for sin and which just gives us a warm unlock are equally warm, ok, they're not there, they're not there, they're not there, they're not there, they're not there there, they're there, so we get a 1 here, they're not going to be here, so we get a zero, they're here, so we get a woman everywhere else, we're going to get a zero zero zero zero zero, okay, so. what we have calculated, well, we are almost there, since we calculated the first term that is here, the second term that is here, the term that is here and the fourth term that is here and what we need to do is what we need o all of them terms together, okay, don't forget that an order only gives us zero, let's call this function s, sorry, okay, that's a function f, so function F is an evaluation of 1/2 of three or zmk which has four. the entries that are defined here here here and here and they are all just due to zero when everything is zero well we have a woman here so no matter what everything else is we will get a 1 we have a 1 here so no matter what else we're going to get a 1 we have a zero here zero but we have a 1 here so we're going to get a 1 okay, we have a 0 here 0 here 0 sorry 0 is 0 is 0 and a 0 here, so we end up with a 0 and we have a 0 here but we have a 1 so we end up with a 1 to 1 and we have a 0 but we have a 1 here so we end up with a 1 ok and you can see there's more here going down so 0 when a 1 is going to be 1 it doesn't matter if it is 0 because it is ordered with a 1 that will give us a 1 and then the end is the first term or the 1 because there is one that is a 0 that is a 1, so we will always get one here and what we just built here is the truth table for the boolean expression that we call F and the output depends on what the input values ​​are, so when a is 0 B is 0 and C are 0 when are all these specific values ​​this function will give us a value of 1 well and when the values ​​are 0 1 and 1 0 for a 1 for B 1 I predict that the output will be what is this is one two three - the fourth down will be zero i so what a tree table gives us is that it gives us all the possible outputs that we could get from a particular expression considering or under the condition of all the possible inputs that we could have, okay guys, and once again, I'm Jonathan Lambert from National College of Ireland mathematics support and development service and I hope this video has been in some way informative. thank you