Greetings Victor, I'll try again. The flaw in your logic is as follows: You show (in the second half of the first sentence of your proof, and explicitly stated in the second sentence) that X and Y are not *both* true. You incorrectly state (by using the word "since" at the end of the first line of the proof) that this implies V(X,Y) is false. That is not correct. As we have agreed, V(F,F)=V(F,T)=T. Neither of these cases requires X=Y=T.
Below, I respond to your previous reply. On 17 November 2011 22:48, Prof. Victor Li <[email protected]> wrote: > in the following discussion we shall use X : = [p] to mean > "X represents p" where p is a proposition. For example, > X : = [P(B) = 1] means "X represents P(B) = 1". >> >> My argument again is: >> a) For some queues in G, (P(B)=1)=F. >> b) if X=F then V(X,F)=V(X,T)=T. >> c) Hence, there exists a queue in G, such that V(P(B)=1, P(S)=0) = T. >> d) Hence, it is not true that for all queues in G, V(P(B)=1, P(S)=0) = >> F. >> >> Please tell me which step you disagree with. > >To avoid any confusion, let us follow the conventions below. >i) To determine the value >of V(X, Y), we must first specify X and Y. >ii) In any expression of the form V(A, B) = C, >A, B, and C must be truth values, rather than propositional variables. >For example, V(X, F) = F is not allowed because X is a propositional >variable. The second proposal gets in the way of communication. See (*) below. > In your argument, suppose that you mean > in a), b) , and c), X : = [P(B) = 1] and X = F; > in b), Y: = [P(S) = 0] and Y = T or F; > in c), for X : = [P(B) = 1] and Y: = [P(S) = 0]. V(F, T) = T and > V(F, F) = T; > in d) X : = [P(B) = 1], Y: = [P(S) = 0], X = T, Y = F and > V(T, F) = F. OK. > You say that for X and Y as specified in d), > V(T, F) = F is not true for all the queues in G because > for X and Y as specified in c) V(F, T) = V(F, F) = T. No. V(T,F)=F is always true. This is (*), the point it is not convenient to express the ideas we are discussing without using propositional variables. What I am saying is that V(P(B)=1, Y)=F is not (always) true, because P(B)=1 is not (always) T. This statement has nothing to do with the value of V(T,F). > Does this mean that you are using the queue specified > in c) as a counterexample to > V(T, F) = F where X : = [P(B) = 1], Y: = [P(S) = 0]? No. Re-read what you wrote. The expression "V(T, F) = F where X : = [P(B) = 1], Y: = [P(S) = 0]" is not useful, because X and Y are never used. I am not saying anything at all about the truth or otherwise of V(T,F). I am talking about the truth of V(X,Y), which, in this case, is V(F, Y) [either V(F,T) or V(F,F), if you want to avoid propositional variables]. > If the answer is yes, then such a counterexample is invalid > as explained below. > > Implications (such as V(T, F) = F with X and Y we specified) > are typically used to express mathematical statements. > If one says "there is a counterexample to an implication" then the > counterexample must satisfy the assumption of the implication for > the counterexample to make sense. If an example does not > satisfy the assumption of an implication, then the example fails to be > a counterexample to the implication, and hence does not make > the implication invalid. It seems that you are using a false > counterexample to argue against an implication. I am not saying the queue is a counter-example to the implication. I am saying it is a counter-example to the statement "For all queues, the implication is false". If there is one queue for which the implication is true, it is a valid counter-example to that statement. The step from (c) to (d) is the fourth equivalence listed at <http://en.wikipedia.org/w/index.php?title=Quantification&oldid=457601452#Equivalent_Expressions>. > If you are not using the > queue as a counterexample, > then your argument is irrelevant to V(T, F) = F > with X : = [P(B) = 1], Y: = [P(S) = 0]. Yes, V(T,F)=F is irrelevant to what I wrote. Please respond to what I wrote. > Return to your conclusion given in d). > You say that it is not true for all queues in G, V(T, F) = F > with X : = [P(B) = 1], Y: = [P(S) = 0]. > Based on our explanation above, > your conclusion appears to be incorrect. The fact is, > there are queues in G that do not satisfy the assumption > X = T with X : = [P(B) = 1] of V(T, F) = F > where Y: = [P(S) = 0] with Y = F. This does not necessarily > mean that V(T, F) = F with X : = [P(B) = 1], Y: = [P(S) = 0] > is incorrect. It does not mean that the statement "V(P(B)=1, P(S)=0) = F" is necessarily incorrect. It does mean that the statement "For all queues, V(P(B)=1, P(S)=0) = F" (the first statement in your proof) is incorrect. > Suppose that one wants to > use a theorem to solve a problem. But the > problem does not satisfy the assumption of the theorem. > Does this mean that the theorem is flawed? Clearly not. > The theorem is still true. Are you suggesting that "P(B)=1" is a hypothesis of your Theorem 1? It is not listed in the statement of the theorem. If not, what exact assumption does my argument not satisfy? In your reply, please do not tell me that I'm trying to deny that "true implies false" is false. Cheers, Lachlan -- Lachlan Andrew Centre for Advanced Internet Architectures (CAIA) Swinburne University of Technology, Melbourne, Australia <http://caia.swin.edu.au/cv/landrew> Ph +61 3 9214 4837 _______________________________________________ IEEE Communications Society Tech. Committee on Computer Communications (TCCC) - for discussions on computer networking and communication. [email protected] https://lists.cs.columbia.edu/cucslists/listinfo/tccc
