What's more important than proving this thing is you gave a proper explanation of what a logical equivalent are🤦♂️ In my University the professor was just like read it this are formulas🤣
The answer for home work problem in Biconditonal Note: Symbols used ('^' and),('√' or),('->' conditional) and ('' Biconditonal) Q] ~(pq) = p~q Here is the solution ~(pq)= ~{(p->q) ^ (q->p)} Using demorgans law in RHS ~(p->q) √ ~(q->p) Using 5th stmt in conditional equivalence (p^~q) √ (q^~p) Let s=p , r=~q So, (s^r) √ (~s^~r) Using 3rd stmt in Biconditonal equivalence s r This is equal to "p~q" I think im correct Thank you
@@royalcanon7433 Hey buddy i think you missed this video on bi- conditional property in the playlist, here is the link of video below ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-ehKd3KmIRSw.html Any way lemme help you out with simple example let us consider this example P: is a polygon with 4 equal sides Q: is a square So, if P is true and Q is true then the proposition is true If P is true and Q is false or P is false and Q is true then the proposition is false This seems a tricky one if P is false (not a polygon) and Q is false(not a square) This makes sense bcoz its not a square since its not a polygon I hope it helps ,Good luck 🎉
@@hariszaib2728 As mentioned here p^1 =p, so the 4th step can be written as (p^1)\/(p^q). Then by taking p as common by distributive law, we get p^(1 \/q)
Negation(p implies q) equivalence p and negation of q Using (p implies q) equivalence to negation of p or q Negation ( negation of p or q) equivalence p and negation of q Then use de Morgan rule P and negation of q is equivalence to p and negation of q
for absorption law (a), another way to think is that either p or p and q needs to be true for the expression to be true. Hence if p is false, the expression has to be false and if p is true the expression has to be true. So its equivalent to p. Similar logic holds for (b)
for the 5th : we have ¬( p -> q ) = p ^ ¬ q ; ................1 we know from the 1st proof that : ( p -> q ) = ¬ p ν q , therefore substituting this same value to : [ ¬( p -> q ) ] we get : ¬( ¬ p v q ) = p ^ ¬ q ; ..............................2 Now by DeMorgan's Law : ¬( p v q ) = ¬ p ^ ¬ q by applying demorgan's law and solving the 2nd equation we'll get : ( p ^ ¬ q ) = ( p ^ ¬ q ) hence therefore, LHS = RHS
Last question. NOT (p biconditional q) is equivalent to (p biconditional NOT q) NOT (p biconditional q) = NOT ((NOT p AND NOT q ) OR (p AND q)) [Biconditional into implication into combination of NOT, AND, OR] = (NOT p OR NOT q) AND (p OR q) [DeMorgan's Law] = (NOT p OR NOT q) AND (NOT(NOT q) OR p) [Double Negation Law] = (p implies NOT q) AND ( NOT q implies p) [Implication] = p biconditional NOT q [Biconditional]
Thank you Neso Academy, detailed explanation, but I have a very confusing question and quiet difficult to break it down, don't know if you can help out
For the Homework (5): is -(p->q) = p and -q are equivalent because if you break it down like following:" -p implies - q" makes that statement True while "p and -q" makes the statement TRUE. Because they both have true values makes the statement true and equivalent. Is that why? Can someone explain to me or check if I'm the right track? Thank you in advance!
You have applied theoretical knowledge of the understanding of the equivalence, I guess it's correct!! The more simpler way that I used is to use De Morgan's laws that sir initially explained to prove it and it becomes just a three liner proof! Hoping that helped... Welcome in advance
now you've said that "-p" implies "-q", and that's true so from there we can agree that "q" implies "p" which is also correct. but the problem is that we can't return back and say that "-q" and "p" are equivalent, the equivalent here is p and "q'' . the homework itself for me is not logical WHY, coze we have -(pq) which is also (pq) but not (p-q). let's make things even more simple, we have "q" and "p", both are equivalent then we say that "p" and "-q" are also equivalent which make no sense like if "a" is "a" then we say "a" is "-a which stands here for another alphabet different from a" and that's not so true.
For the homework #5 I got, -(p -> q) == -(-p v q) {Conditional Law} == - -p^-q {DeMorgans Law} == p^-q {Double Negation Law} Please let me know if this is right or how I did!
For the homework #4 involving bi-conditional statements like the following: "-(p q) = p -q " For this one I broke it like this "(-p -> -q) or (-q -> -p) = (p-> -q) (-q-> p) . Based on the Double negate law, " (-p-> -q) or (-q-> -p) is True as well as (p->-q) (-q-> p) which are True because no matter what if p is false it doesn't matter if q is False or True, p->q will always be TRUE. This is why they are equivalent. Any one want to give suggestions if I'm the right track here?
pls check if i am correct with 5th homework task: NOT(p=>q) p and NOTq We can transform it as following: (we can do a double negation of both sides) NOT(NOT(p=>q)) NOT(p and NOTq) (p => q) (NOTp or q) (right side is the same as 1st conditional statement) please correct me!
5:37 😂 4th line is same place where you started to prove formula . By the way I like Boolean notation more , it's easier because we are familiar with + , .
9:45 my dummy self would have put a double negation on the p. But I guess it is commutive(you can switch the p's and q's) Would it be wrong to do that? Is it only logically equivalent if you give it no one that isn't already negated?
thank you very much dude, I am just learning this for myself as I have graduated from school long time ago. but bruh! wat in the multiverse is this shoot! Aliens laugh at us with all these nonsensical convention we brought. I am learning and laughing at this!
