Discrete math help

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Levanion

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Hello, I'm doing some practice problems on zybooks and I'm having trouble understanding why I'm getting the same problems wrong. I provided a screenshot of the two I keep getting wrong and I really don't want to guess on them to get it right, or use the logic that since I know what the wrong answer is, I should just do the opposite of what I think. Can someone please explain the process or what I may be missing with these two truth table's truth values/how you get to that conclusion?
 

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Logic is a course that I taught most often. That said, I must tell you that I do not follow your question.
So lets get it straight as to what it means exactly.
\(\begin{array}{*{20}{c}}
P&Q&R&|&{P \leftrightarrow (Q \vee R)} \\
\hline
F&T&F&|& {\color{red}T} \\
\hline
F&F&T&|&{\color{red}T}
\end{array}\)
Please tell us if the above is correct. Tell us if you put the two red T's and they are incorrect and you would like to understand why.
I will wait to help until you respond.
 
Levanion said:
Hello, I'm doing some practice problems on zybooks and I'm having trouble understanding why I'm getting the same problems wrong. I provided a screenshot of the two I keep getting wrong and I really don't want to guess on them to get it right, or use the logic that since I know what the wrong answer is, I should just do the opposite of what I think. Can someone please explain the process or what I may be missing with these two truth table's truth values/how you get to that conclusion?
Click to expand...
It will be very helpful if you can tell us why you gave the answers you did for those two parts. We can help you better by identifying your misunderstanding than by just telling you again what your book will already have told you, and guessing what you may be doing wrong.
 
Not to hijack the thread, but I am dealing with the same problem; and I suspect that user "Levanion" might relate to my confusion on the matter. Specifically, how the statement "~p --> q, r:True" (that is, "not p implies q is true") makes any ordinary sense. I don't see how the negation of the hypothesis, p, must have any correlation with the truth/untruth of the statement as a whole. This sort of logic does not seem to map onto my colloquial, or non-mathematic, notions of logic.
 
love/hatewithMath said:
Not to hijack the thread, but I am dealing with the same problem; and I suspect that user "Levanion" might relate to my confusion on the matter. Specifically, how the statement "~p --> q, r:True" (that is, "not p implies q is true") makes any ordinary sense. I don't see how the negation of the hypothesis, p, must have any correlation with the truth/untruth of the statement as a whole. This sort of logic does not seem to map onto my colloquial, or non-mathematic, notions of logic.
Click to expand...
Do you understand that the following table defines the truth values of implications, if P then Q?
\(\begin{array}{*{20}{c}} p&q&{p \to q} \\ \hline T&T&T \\ T&F&F \\ F&T&T \\ F&F&T \end{array}\)
__\(\begin{array}{*{20}{c}} { \neg p}&q&{\neg p \to q} \\ \hline F&T&T \\ F&F&T \\ T&T&T \\ T&F&F \end{array}\)

In both tables can you see that
TRUE statements are implied by any statement.
FALSE statements imply any statement ?
 
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I see that classification of the whole statement as FALSE results only when p, or ~p, is true, and q is false; and that every other combination on the table results in a TRUE classification. But I don't see why this is so.

One professor gives the following example:
truth values.png
​
He goes on to say that since the hypothesis is false, that the whole statement is necessarily true, regardless of what B is. How can that be so if, for example, B happens to be an empty set as well? I can't put my finger on it; but it feels like "implies" is used in an extraordinary way in this context.
 
Did you read my post?
TRUE statements are implied by any statement.
FALSE statements imply any statement ?
Suppose that X is any statement whatsoever. then If X then 2+2=4 is true.
If 2+2=5 then X is a true statement.
 
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love/hatewithMath said:
Not to hijack the thread, but I am dealing with the same problem; and I suspect that user "Levanion" might relate to my confusion on the matter. Specifically, how the statement "~p --> q, r:True" (that is, "not p implies q is true") makes any ordinary sense. I don't see how the negation of the hypothesis, p, must have any correlation with the truth/untruth of the statement as a whole. This sort of logic does not seem to map onto my colloquial, or non-mathematic, notions of logic.
Click to expand...

I'm not sure you mean exactly what you wrote; "~p --> q" doesn't claim that p is not true, only that if p is not true, then q is true, which makes perfectly good sense. I think what you mean to ask is, why, if p is true, is that conditional statement considered true? That is, if the hypothesis is false, why is the conditional statement called true?

This is a common question, to which there are several answers, some in everyday thought and some in terms of formal logic. Ultimately, though, it is a matter of definition. We choose to define the conditional statement this way within mathematics, because it makes certain aspects of logic work. And we have to make some definition, unless we want to say that F-->T is just undefined, or some third logic state.
  • Suppose I say, "If I win this game, I'll give you $100", and I lose the game. I give you nothing. Did I lie? Or, I give you $100 anyway. Did I lie now? No, because I made no promise about what I would do if I didn't win.
  • What we are doing there is following the idea of "innocent until proven guilty": a statement can't be called false without evidence. So we call it true in this case, because we need a two-valued logic.
  • In particular, one application of conditional statements is in analyzing logical arguments, where we might say, "If [premise 1] and [premise 2], then [conclusion]". We consider the argument valid if the conditional statement is true regardless of the truth values of the premises. But if we called a conditional statement false when the condition is false, then valid arguments would be called invalid. We must define the conditional as we do in order to talk about valid arguments in this way.
I'd also like to add that the word "implies" is misleading, as that tends to imply a causal connection, or at least a necessary relationship. A conditional statement can be given just about one situation: "If my car is blue, then I am rich" -- it is not blue, and I am not rich, so this is logically true, but it doesn't mean that if my car were blue, I would be rich. It's just a logical statement, and nothing more.

I expressed these ideas in more detail in my blog here, where I reviewed old questions about this topic.

(By the way, you did hijack a thread that has absolutely nothing to do with your question. There's no reason not to have posted it as a new thread.)
 
I will think on this. Dr. Peterson: I don't see how my contribution is impertinent to this thread, given that the OP posted a screenshot of a truth table for conditional propositions, the function of which I thought was more or less the subject of discussion since I arrived; but my failure to recognize the lack of relevance of my question to the thread should perhaps be unsurprising, as I have yet to find much like what ordinarily passes for meaningful and actionable sense in subsequent replies.

In reply to pka, I read your post. You write English in incomplete sentences, dispensing with grammatical convention, yielding some product which does not clearly connect with the tables to which you refer; thereby making it all somewhat less than perfectly intelligible. I do appreciate your effort.
pka said:
FALSE statements imply any statement ?
Click to expand...
 
love/hatewithMath said:
I don't see how my contribution is impertinent to this thread, given that the OP posted a screenshot of a truth table for conditional propositions, the function of which I thought was more or less the subject of discussion since I arrived
Click to expand...
The original question was about a biconditional, not a conditional; and the wrong T answers have nothing to do with your question about why T is correct for a conditional.
 
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