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Fax initial termination

okay so to finish up the the total correctness proof for the same algorithm so in the last video I did the partial correctness proving that it it always guarantees the post condition in previous videos I've kind of explained why proving those three properties all the variant that belongs does have a natural is that if it equals zero it um but it terminates and that it reduces every iteration why if you can guarantee those three properties then it proves that the loop will terminate um so now we're just going to try and come up with a with a variant and then kind of prove those three properties so first of all the property that always think about festival is just what what is always reducing and will equal zero so like if a equals zero over here the loop doesn't terminate if y equals zero the loop doesn't necessarily terminate um so what I might try here is that like a times y sorry a minus y it's just kind of a just like a intuitive guess because if a is equal to Y then a minus y will be zero because if a is equal to Y then a minus y is equal to a minus a which is equal to zero I so therefore if a minus y is equal to zero then a is equal to Y in which case this loop terminates so that's pretty much my might variant explanations at least at the moment I'm gonna ask one about this but at the moment they include kind of like a lot of normal language sentences like that just do like proving it so you can say that you know like if a minus y equals zero than a - y the condition is B therefore a equals y is the equivalent to not be therefore if a minus y equals zero the loop terminates he's just kind of like the the logical explanation I go through for that but we can all kind of see that that if a minus y is equal to zero that the loop will terminate so there's our first property kind of explained the second one is that it needs to belong to the set of natural numbers sir it's initialized to zero and don't we ever gets incremented by one so again just kind of plain woods description that a belongs to the set of natural numbers and why here is stated as greater than or equal to zero so therefore if it's a natural number well you could even kind of like a list as a precondition because I guess does this loop walk if Y is equal to 1.5 and the answer is no because it will skip pasta J will never equal Y this will go forever and that's actually not listed as precondition it does still meet the requirement that's greater than zero so you can actually just kind of like right on that you need to assume why is it in to jail or something like that because a that is actually a requirement that is algorithm to work but basically if a is a something that's said natural numbers and Y is belongs to the set of natural numbers then say a starts at zero and Y is greater than or equal to zero and then what we actually get is that maybe this doesn't work because let's say like Y is 5 right well then 0 minus 5 gives us negative 5 which does not belong to the set of natural numbers negative numbers in all natural numbers so that's kind of whoops we screwed our Varian but that's fine so our proof that this was the variant we're just trying to find out what a variant is it was wrong the first time around but probably on the right track so let's try Y minus a instead rights this has the exact same proof for y when that equals 0 they're the same because when y minus a equals 0 then y equals a and what Michael's a the loop terminates so we've got our exact same proof from before now we know the equal 0 and Y is greater than 0 so we know that y must be greater than or equal to and we know that they both belong to the set of natural numbers and so from there we can kind of see that since a starts a zero if Wisel say T equals zero then the loop terminates otherwise Y minus a for any condition would actually run through the loop will always belong to the set of natural numbers and again I'm not sure if these kind of plain woods descriptions are enough for the assignment this is what I want to double check but at the moment I'm just kind of looking through it and the final condition is just that what have we done so far we've done that when they equals or the loop terminates we've done that they belong to the set of natural numbers oh that it decreases every time okay sir yeah here we've got now we would have run into the same issue here if we hadn't called it Elya but so if a variant is equal to Y minus a then every iteration you've got a plus one so if we kind of say and I think the way they did in one of the things was that yeah they've in said the almost an accession now that Y minus a is equal to the original value of our variant which they called like capital I but I think that's because there was a little I so but let's just use it anyway so Y minus a is equal to capital I whatever it happens to be and then we've got the Z equals a plus X that doesn't affect it equals a plus one which means Y minus a plus one is no longer equal to AI but that is in fact greater than I and that's just to kind of like basic math you could substitute this back in so Y minus a plus one is greater than Y minus a because like math axioms X is less than X plus one little shed so this statement is true and since this was the initial value and this is the like adjusted value we know if a sudden that Y minus a plus one oops sorry I forgot the brackets when you substitute that in it's y minus a plus one no y minus a plus one now which means when you expand that sort of totally flipped this around the wrong way we have y minus a minus one which will always be less than Y minus a for the same reason so we've proof that the new value will always be less than the original value and thus we've proved that out there in Y minus a is decreasing every iteration and so again I'm not sure how formal all that needs to be but that's kind of the general process you go through to prove the variant and then between those two the marshal proof correctness and the proof that the variant satisfies those three conditions you have the total correctness