I'm teaching my little sister propositional logic per her request.  I was trying to explain to her why $A \implies B$ holds whenever $A$ is false, and I didn't succeed with that.  I referred her to the definition: $A \implies B$ is true if whenever $A$ true, $B$ is true. If $A$ is false, then it is "never" true. If we want to test whether the implication is true, we need to check if $B$ follows when $A$ is true, but $A$ is never true, so we need not check anything, hence the implication is true.   Is my reasoning even correct? I forgot how to do these stuff anyway.   Please help me explain this idea. It would be nice if you can give an intuitive way to think about it as well.
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I am studying entailment in classical first-order logic.  The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is: $$\begin{array}{|c|c|c|} \hline p&amp;q&amp;p\Rightarrow q\\ \hline T&amp;T&amp;T\\ T&amp;F&amp;F\\ F&amp;T&amp;T\\ F&amp;F&amp;T\\\hline \end{array}$$  I 'get' lines 1, 2, and 3, but I do not understand line 4.      Why is the statement $(p \Rightarrow q)$ True if both p and q are False?  We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).  Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.       Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question  And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
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Am unsure (leaning more towards no): true P and true C is valid, false P and false C is valid, True P and false C is invalid, how about false P and true C?  P= Premises; C= Conclusion.  People who says.. Yes, it is valid, I would like to hear some answers for   How can we derive the truth out of falsehood?  If we can, then True P - False C is valid too (tell me why it is invalid?).
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I have a question for my exam and I find it hard to understand.  I have to prove that the following formula is logically valid:    The professor told me to "push" all the symbols inside the brackets, and use the deduction theorem.  But I don't know how to do it, because I can't find the identities to push the "exist" symbol inside the brackets.  Your help is appriciated, thank you.  Alan
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Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$? I started with the hypotheses $(p \implies q) \implies p$ and $\sim p$. However, from these hypotheses I did not get the desired contradiction to solve the question.  Thanks sb45
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Does $\exists$ distribute over an implication? ie. Is $\exists x \in \mathbb{R}, (p(x) \rightarrow q(x))$ logically equivalent to $(\exists x \in \mathbb{R}, p(x)) \rightarrow (\exists x \in \mathbb{R}, q(x))$. If so, can you give an example of $p(x)$ and $q(x)$ to demonstrate?  Thanks very much for any help in advance.
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I am not able to understand implication well. I can see If in $P \implies Q$, if Pis true and Q is false then implication is false. But i cannot understand other cases where it is taken as true. Is this just a convention ? or some underlying reasoning behind it  Thanks
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Let a statement P is "X is true if and only if Y is true". What is the negation of P? I am little confused. It seems that digital equivalent of this statement is P = X and Y. Hence negation of P is (not X) or (not Y) i.e. Either X or Y is false. Am I right guys?
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I was pondering a bit more about  question regarding being able to &quot;omit&quot; the base case in a proof by strong induction due to vacuous truth. The post states:  Strong induction proves a sequence of statements $P(0), P(1), …$ by proving the implication &quot;If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true.&quot; for every nonnegative integer $n$. There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously.  However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds.
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On  in XI Free Models and Logic Programming of Ebbinghaus' Mathematical Logic:  In general, the following statement is false: (*)  If $\Phi \models \exists x \phi$  then there is a term $t$  with $\Phi \models \phi \frac{t}{x}$ We  get  a  counterexample for  $S  = \{R\}$  with unary $R$,  $\Phi = \{\exists x Rx\}$,  and  $\phi  =  Rx$.  In the counterexample, why if $\{\exists x Rx\} \models \exists x R x $, then there is no term $t$   with $\{\exists x Rx\} \models R t $? Thanks.
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I'm currently studying logic and I'm confused about what is meant by the truth or falsity of statements within an argument. From Daniel J. Velleman's, How to Prove It, pg.9:  &quot;Although we have no guarantee the conclusion is true, it can only be false if at least one of the premises is false ... We will say an argument is valid if the premises cannot all be true without the conclusion being true as well.&quot;  Having then read a  about the validity of an argument, I am told the following:  &quot;The following argument is of the same logical form but with false premises and a false conclusion, and it is equally valid: All cups are green.  Socrates is a cup. Therefore, Socrates is green.&quot;   I feel that there is some distinction being made here between the terms &quot;truth&quot; and &quot;false&quot; in the two instances that I'm not quite grasping. In the wikipedia article I believe &quot;false&quot; refers to the general truth of each individual statement i.e. we know all cups are not green in reality so this is a &quot;false&quot; statement. However in the Velleman book I am not sure what he means by the &quot;truth&quot; or &quot;falsity&quot; of the premise and conclusion. In sentential logic what is meant by the &quot;truth&quot; or &quot;falsity&quot; of a statement – the &quot;empirical truth&quot; of the statement, or does it refer to some other notion of &quot;truth&quot; and &quot;falsity&quot;?
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I understand that  (1) "$A$ if and only if $B$" ($A\iff B)$  means that  (2) "$A$ implies $B$ and $B$ implies $A$" $(A\implies B)\land (B\implies A)$.  The phrase "$A$ if and only if $B$" sounds as though something applies exclusively or precisely in case something else is true, i.e., not only does $B$ make $A$ true, but $B$ is the only thing that makes $A$ true. If this is the case, then how exactly does this interpretation relate to two things implying one another?    For the Bounty: (1) I think that a natural interpretation of the statement $A$ if and only if $B$ is that $A$ is true precisely in case $B$ is true, i.e., if $B$ is true, then $A$ is true, and if $B$ is false, then $A$ is false. Is this correct? And if so, how does this interpretation mean that $A$ and $B$ imply one another and also have the same truth value?    (2) Related to (1), the (true) statement $\forall x\in\mathbb R:2x=5\iff x={5\over 2}$ means that for any replacement of the variable, the two statements have the same truth value. How does two statements having the same truth value in all cases mean that they imply each other?  (3) Most importantly, I need exposition. Am I missing anything obvious, or is this a simple thing?
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True or False  $A - C = B - C $ if and only if $A \cup C = B \cup C$   I am in an introduction to proofs class.  I think this is a true statement. I have began the proof and realize I have to do this both directions.  I am stuck on the second part of the proof.  Assume $A - C = B - C$ Let $x \in A$ then $x \in A$ but $x\notin C$ By the assumption $x \in B - C$ then $x \in B$ and $x\notin C$  If $x \in A$ then $x \in A \cup C$.  If $x\in B$ then $x \in B \cup C$. Therefore,  $A \cup C = B\cup C$.   Assume $A \cup C = B \cup C$. Let $x \in A \cup C$. Then $x \in A$ or $x\in C$.  If $x \in A$ then $x \in A - C$  If $x\notin A$ then $x \in C$.   This is where I get stuck... I know I am trying to show $A-C = B-C$ for the second part but I don't know how to get there... Am I on the right track?
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I'm having trouble solving a Fitch Style Proof and I was hoping someone would be able to help me.  Premises: p ∧ q  Goal: q ∨ r  Would be an assumption (p ∧ q => q ∨ r) in first step a correct one?   Then, q ∨ r Implication Elimination: 2, 1  Why is it not correct?  Thank you very much for any help!
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I am taking a real analysis course. I have the following statement:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a &lt;  \epsilon$, for all $\epsilon &gt; 0$, then $a = 0$   I know how to prove it (using contradiction). Now, I was wondering, how I prove almost the same thing:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a \leq \epsilon$ for all $\epsilon &gt; 0$, then $a = 0$   Can someone show me how to do it? I am stuck with the equality to show a contradiction.  Note: this is not homework, just for my own knowledge.
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I'm not quite sure that I really understand WHY I need to use implication for universal quantification, and conjunction for existential quantification.  Let $F$ be the domain of fruits and  $$A(x) : \text{is an apple}$$  $$D(x) : \text{is delicious}$$  Let's say: $$\forall{x} \in F, A(x) \implies D(x)$$ Is correct and means all apples are delicous.   Whereas, $$\forall{x} \in F, A(x) \land D(x)$$ is incorrect because this would be saying that all fruits are apples and delicious which is wrong.  But when it comes to the existential quantifier: $$\exists{x} \in F, A(x) \land D(x)$$ Is correct and means there is some apple that is delicious.  Also, $$\exists{x} \in F, A(x) \implies D(x)$$ Is incorrect, but I cannot tell why. To me it says there is some fruit that if it is an apple, it is delicious.  I cannot tell the difference in this case, and why the second case is incorrect?
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How is the statement $\bot \to \top$ true? If, for example, I take the statement  $P = 2$ is an odd number $Q = 5$ is an odd number Then $P$ is false and $Q$ is true.  Clearly, false implies true translates to  if $2$ is an odd number then 5 is an odd number  which means that  if $2$ is not an odd number (i.e. even) then $5$ is not an odd number  which is clearly false as $2$ is indeed an even number. Please help me to find what I’m missing out in this. Thanks for your help!
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I am reading Velleman's "How to Prove it", and in one example he proves  $\neg \exists x P(x)  \implies \forall x \neg P(x)$    in the following way:    Suppose $\neg \exists x P(x)$. Let x be arbitrary. Suppose $P(x)$. Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x)  \implies \forall x \neg P(x)$    What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?    Edit: I deleted my first example since it was clearly not representative for the problem.  Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does.