@jay_csai
C. A is knave, B is knight D. A is knight, B is knight E. Can't say; any of the possibilities could be true
@rodrickngonyoku5042
I am so confused š¤¦.
@KiranShirke3y7
Everyone can confuse 1st time. I can watch the video and then I am also confused and i stop the study and say "faltu question".... But next day I go to study and say yourself "1bar try karte hain". And finally I know how to solve this questions and get relaxed...šš
@MDgaming789
Thanks... Now Finally I can Complete First Question of My Assignment. And Move On
@devukun8358
you know you're fcked when you brain stops functioning mid way .... thank you for the lecture sir :)
@timothywaters8249
Hopefully you are still reading comments... I was hoping to see a few more examples with rigorous proofs using your method along with truth tables. The process of how you solve these logic problems is important as they can get cumbersome with many conditions, but knowing <how> to solve them rigorously would help us tremendously. Thank you, love your courses!
@PowerStar004
C: If A's statement is true, than B's statment is false, which is contradictory. Thus, A's statement is false, inturn making B's statement true. A - Knave, B - Knight D: The intended solution is likey that A being a Knave would make the statement true, hence they must be a Knight, which means the statement must be true, thus B must also be a Knight. However, the question writer appears to have forgotten about exclusive or, where if both statements are true, then the or statement itself is false. Thus, a Knave could say "I am a Knave or B is a Knight", so long as the or is exclusive and B is a Knight. Without the assumption that the "or" is specifically INCLUSIVE or, there is more than one possible answer. A - Either, B - Knight E: There is no way to tell. Both A and B could be either a Knight or a Knave. A - Either, B - Either
@nnaammuuss
(a) A is a knight => B is a knight => A is a knave (the opposite type). A contradiction. Therefore, A is a knave. Therefore, B is a knave (which works out, as they are then the same type). (b) A is a knave=>both are knights=>A is a knight. A contradiction. So, A is a knight. => B is a knave. ā
@nathWSD
for E the solution goes forth:
either A is a knight and B is a knight
A is a knave and B is a knave
A is a knight and B is a knave
A is a knave and B is knight
those are the possiblities to be tested at once
@JatinKumar-w7f
the opposite of "at least one of us is knave " can also be both are knave, and if both are knave the second proposition is true. here we can come up with different inference that both are knave
@kvsspkaushik2307
IS it only my brain which is dumb or everyone else
@marco_robert
8:45 The expression could be "not p or not q" too. It eventually means at least one is knave. Right?
@saivamsi854
Sir please post correct answers for home work problems
@xuliganxuligan-w2w
my brain: "ouh yeah eazy but how" lol xaxaaxa
@subratadutta7710
Explanation šš„
@sadmanmohammadnasif8830
c) A is a knave, B is either Knight or Knave.
d) Either (A=Knight and B=Knight), Or (A=Knave and B=Knave).
e) A and B can be either Knight or Knave. Whatever they are, (e) is consistent.
Reasoning:
c) If P=F, then (P^Q) has to be F. That concludes for any value (T or F) of Q , (P^Q) is F. So P=F, Q=T or F.
d) If P=T, then (Ā¬PvQ) has to be T which is only possible if Q=T. So, P=T and Q=T.
If P=F then (Ā¬PvQ) has to be F which is only possible if Q=F. So, P=F and Q=F
e) If P=T, then P has to be a Knight, so P=T.
If P=F, then P has to be knave, so P=F
If Q=T, then Q has to be a Knight, so Q=T.
If Q=F, then Q has to be knave, so Q=F. Therefore, P and Q can be anything for (e) to be consistent.
@yahyairfan1159
c)A.knave,B:knight d.both are either knave or knight e.both are either knave or knight
@madaragimhani8747
Great explanation ā¤
@meetjoshi4521
For answer D:
A cannot be a knave . Why?
Ans: if he is a knave then his statement will always become true ("I am a knave" becomes true and "true or B is a knight" will always be true)which is contradiction.
Hence A will always be knight. if so then he must always say true .TO make his comment true B has to be Knight.
So final answer : A=Knight B=Knight.
@vasubhatt6160
ā yes I can explain just consider for case C) If you assume A-knight then according to him B is also knight on the other hand B is saying A-knave so it's contradiction so A can't be true and when you assume B-knight then that matches with the sayings of both as B is saying A-knave and A is saying both of us are knight which is false so RESULT A-knave B-knight case D)If you assume A-knight so according to him either I am knave OR B-knight but as we assumed A-knight he cant be knave so B-knight but as some of the people in comments are saying both of them can be knave which is not true, if both of them are knave then the statement of A must be false which ic ~P OR Q but itwill never be false because of NOT P(NOT*FALSE=TRUE) so RESULT A-KNIGHT B-KNIGHT case e) is a funny case because you can't conclude anything, because it could be anything 1)A-KNIGHT B-KNIGHT 2)A-KNIGHT B-KNAVE 3)A-KNAVE B-KNIGHT 4)A-KNAVE B-KNAVE here they are just saying about themselves and not for the other person so either they are saying the truth or false, GOD knows:")