as we discussed in the last lesson really the only way to get good at simplification of boolean expressions practice let's show you just two more examples in this lesson just so we can grab a little bit more practice when it comes to these expressions in fact let's try an expression oh i don't know this one may be a little bit much to look at first but hopefully it'll make sense when the deeper we get into it so let's say i've got a and b bar and c ored with a bar and b bar and c ord with let's see a bar and b bar and c bar ord with a bar whoops now let's make that a and b bar and c bar hopefully that all fit in the camera now anything that you see in terms of a red flag and remember our red flags mean something's being combined with itself something's being combined with its inverse or something's being combined with a constant the only constants we have available to us 0 and 1. so don't see it i don't see anything right away what we can see though is that b-bar is in every single one of those products so we can use the standard form of the distributive law to pull out a b bar so i can pull out b bar ended with a and c or a bar and c or a bar and c bar or a and c bar okay still no red flags well what else could we massage around what else could we move around when it comes to this expression to make it so that it's possibly a little well so something shows up that we can simplify well if you see these first two terms they have one has a one has a bar they both have c if we pull c out we can see a or a bar let's go ahead and do that for those first two terms so kind of embedding these parentheses a or a bar and then we can do a similar thing for this last term i can pull out c bar and have a bar or a now and in fact you know we've talked about this before where there's usually more than one path to reach a final simplification if i had swapped these two terms now let's see if i had swapped these two terms through the commutative laws so the the this product swapped with this product instead of pulling c out i could have pulled a out of the first two terms and a bar out of the second two terms so really the path we're taking just happens to be the path we're taking now we do have a red bar excuse me a red flag showing up don't we we have something ored with its inverse and something ored with its inverse turns out it's the same thing there's in fact two ways we could do this i could pull a or a bar out of this product and pull a bar or a out of this product pull it out and get a or a bar anded with c or c bar but that's not what i'm going to do what i'm going to do instead is see i've got something being ored with its inverse what happens when you or something with its inverse you get one again here i have something ored with its inverse i get one all right now i have a glaring red flag right there's a constant that has showed up here this constant well anything anded with one always itself so this becomes b bar anded with c or c bar and then we once again have something being ored with its inverse that becomes one this becomes b bar ended with one that's just b bar and so the final simplification wow the final simplification i have started out with one two three four three input products ored together with one four input or gate that's a lot of gates in the end just an inverter that haul can be taken care of with an inverter now there's something else to kind of see here and if you're familiar with this idea of ones and zeros and the patterns of ones and zeros you'll see that yes once we get to this second step here this inside of the parentheses has a unique pattern what it's saying is if a is a 1 and c is a 1 or if a is a 0 and c is a 1 or if a is a 0 and c is a 0 or if a is a 1 and c is a 0 then everything inside the parentheses is going to be one well it turns out there's no other pattern of ones and zeros that a and c could possibly take on so therefore it has to be one this case this case this case this case whatever a and c are it has to be one of these and so exactly one of these products is going to be one and since there's a one inside of this or inside of this four input or gate we're going to have a 1 there we just did it using boolean using boolean identities from here down so there you go let's try one more all right now this next expression well it may not be as obvious as the previous one but let's go for it i've written it down because i can't remember it let's see how about a ended with c word with a ended with d ord with a ended with c bar and d bar ored with a ended with b wow there really isn't any red flag showing up there is there well we do have an a in every one of those products so we have the four products that are going into a four input or gate and a and none of them are inverted because if one of those were inverted we can't pull out an a bar with an a at the same time all four of those a's don't have a bar over top of them so i'm going to go ahead and pull out that a pulling out that a is going to give me c or d or c bar and d bar or b right all right now there are a couple of ways we could do this first of all do you remember we had an expression or excuse me a a property that a ord with a bar b is equal to a or b do you remember that one well we have that kind of here don't we if i look at these two terms right here and i substitute d for a which means d bar is substituted for a bar and then c bar is substituted for b this guy is actually equal to d or c bar so those two products and and d really is a product it's just a product with only one input d or with c bar d bar is the same thing as d or c bar now let's bring down the rest of what's inside the parentheses and also bring down the a so then now we've got this moves to this once we apply this rule right here the other thing that we could do is use the commutative law and the commutative law says i can take these two terms and swap them around all right hopefully you see the red flag at this point we've got something ored with its inverse all right something ordered with its inverse is always one so we get one or d or b of course don't forget our end with a anything in order excuse me anything or'ed with one is one right a one goes into an or gate a one comes out of the or gate so we've got a and one which is equal to a all right well cool there's another way to do this though and this other way is very different than the application of this rule what i could do is start with this guy in fact let me go ahead and bring this one down i'm going to a little line here we're going to bring this guy back down here so we'll have a ended with c or d or c bar d bar or b okay now this may not be obvious to somebody who is just starting out in simplification but it turns out we have an application of de morgan's theorem one of de morgan's theorems forms in this in this expression remember what i've got is there were two versions of de morgan's theorem one said a bar or excuse me a or b the inverse of a or b is equal to a bar ended with b bar right the other form said that the inverse of a and b is equal to a bar or b bar the way that worked was i can distribute the bar to the inputs as long as i change the operation i can distribute the bar to the input as long as i change the operation well look at this guy i could actually apply this backwards right the other way around instead of distributing i could pull the bar out look at this c bar and d bar c bar and e bar that looks like this form right if i go backwards and make it a or b inverted i get a anded with c or d or with c or d inverse the inverse of c or d or b all right now the commutative law says i could also just simply put parentheses around c or d and look at this one step earlier we got something ored with its inverse this is just by an application of de morgan's theorem i've got something ored with its inverse what happens when i or something with its inverse that's going to be equal to 1. what happens when i or 1 with something at equals 1 which equals just a and so our final result no matter what path we took is simply that gosh once again i've got 4 products and one four input or gate combining all of these signals together all of that filtered its way down to just a wire just a is driving whatever as long as a is true what we're looking for this expression right here it's true too all right hopefully the three examples from the last lesson and the two examples from this lesson will give you an idea of how to slowly but surely move your way forward through simplification of these boolean expressions