According to my textbook the implication P --> Q has the following truth table:    I don't understand the last two rows. For example in the last row, how can we determine P --> Q if all we know is that both P and Q is false. Shouldn't the truth value be unknown in that case?
5100
Provided we have this truth table where "$p\implies q$" means "if $p$ then $q$":  $$\begin{array}{|c|c|c|} \hline p&amp;q&amp;p\implies q\\ \hline T&amp;T&amp;T\\ T&amp;F&amp;F\\ F&amp;T&amp;T\\ F&amp;F&amp;T\\\hline \end{array}$$  My understanding is that "$p\implies q$" means "when there is $p$, there is q". The second row in the truth table where $p$ is true and $q$ is false would then contradict "$p\implies q$" because there is no $q$ when $p$ is present.  Why then, does the third row of the truth table not contradict "$p\implies q$"? If $q$ is true when $p$ is false, then $p$ is not a condition of $q$.  I have not taken any logic class so please explain it in layman's terms.       Administrative note. You may experience being directed here even though your question was actually about line 4 of the truth table instead. In that case, see the companion question  And even if your original worry was about line 3, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
102482
Is this proof correct?  p → q, ¬p → r, ¬q → ¬r ⊢ q  1) p → q      premise 2) ¬p → r     premise 3) ¬q → ¬r    premise   4) -q       assumption   5) -p       MT 1,4     6) p      assumption     7) ⊥      ¬e 5,6   8) ⊥        ¬e 2,3 9) q          ¬e 8   Am I going correct? What might be the next step?  I updated my answer.
27891
Can someone provide me with the simplest possible argument for why the material conditional has the truth table it does?   Googling doesn't help at all, providing only flimsy and clumsy arguments. I want the argument to be simple, and yet as convincing as possible.
83736
I just started doing logic and everything is going fine. But we just got to Material Implication and I don't really understand it. From for example $q$, we can get $q \lor \lnot p$ by Disjunction, and from there, using Material Implication, we can get $p \to q$.   The thing I don't understand is how is this valid. Let's say    $p$ is "it's raining", and  $q$ is "the floor is wet",    and $p \to q$ holds. But just because the floor is wet, doesn't mean that it's raining. So how can we conclude that $p \to q$ just from $q$?  I'm sorry if it's confusing.
73109
Does the equivalence $$\neg\neg (P \rightarrow Q) \leftrightarrow (\neg\neg P \rightarrow \neg\neg Q)$$ hold in propositional intuitionistic logic? In propositional classical logic the equivalence holds obviously since $P \leftrightarrow \neg\neg P$. However, the similar-looking implication $\neg\neg (P \vee Q) \rightarrow (\neg\neg P \vee \neg\neg Q)$ is . The validity of the backward implication comes up in  as well.
47685
I know, I know, there are tons of questions on this -- I've read them all, it feels like. I don't understand why $(F \implies F) \equiv T$ and $(F \implies T) \equiv T$.  One of the best examples I saw was showing how if you start out with a false premise like $3=5$ then you can derive all sorts of statements that are true like $8=8$ but also false like $6=10$, hence $F \implies T$ is true but so is $F \implies F$.  But for me examples don't always do it for me because how do I know if the relationship always holds even outside the example? Sometimes examples aren't sufficiently generalized.  Sometimes people say "Well ($p \implies q$) is equivalent to $\lnot p \lor q$ so you can prove it that way!" except we arrived at that representation from the truth table in the first place from disjunctive normal form so the argument is circular and I don't find it convincing.  Sometimes people will use analogies like "Well assume we relabeled those two "vacuous cases" three other ways, $F/F, F/T, T/F$ -- see how the end results make no sense?" Sure but T/T makes no sense to me either so I don't see why this is a good argument. Just because the other three are silly doesn't tell me why T/T is not silly.   Other times I see "Well it's just defined that way because it's useful"... with no examples of how it's indeed useful and why we couldn't make do with some other definition. Then this leads to the inevitable counter-responders who insist it's not mere definition of convenience but a consequence of other rules in the system and so on, adding to the confusion.  So I'm hoping to skip all that: Is there some other way to show without a doubt that $(F \implies q) \equiv T$?
107714
I am trying to understand the proof above, Proposition 7, given by the author, but I got stuck at some points.   My explanation:  Now, suppose for contradiction there is an $\epsilon &gt; 0$, for all $x &gt; 0$ such that  $NOT (x^2 &lt; 2 &lt; (x+ \epsilon)^2)$,  Since $NOT (x^2 &lt; 2 &lt; (x+ \epsilon)^2) \ \ iff \ \ (x^2 \geq 2 \ \ OR  \ \ (x+ \epsilon)^2 \leq 2)$,  We must show that for some $\epsilon&gt;0$ and for all $x &gt; 0$, the statement $x^2 \geq 2 \ \ OR  \ \ (x+ \epsilon)^2 \leq 2$ leads to a contradiction.  I think in the proof above, author tries to get a contradiction using right hand side of $OR$ clause, namely $(x+ \epsilon)^2 \leq 2$. But no contradicition is achieved by him for the left hand side side of $OR$ clause namely, $x^2 \geq 2$.  First Question is, am I correct at this point, i.e. the need for a contradiction for the left hand side side of $OR$ clause?  Second question is, I don't understand how he derived the following sentence in his proof above,     Since $0^2 &lt; 2$, we thus have $\epsilon^2 &lt; 2$, which then implies $(2 \epsilon)^2 &lt; 2$ and indeed $(n \epsilon)^2 &lt; 2$   Why does he feel a need to multiply $\epsilon$ constantly by $n$
5225
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.
109026
Given $x$ occurs free in $P$. If $x$ does not occur free in $Q$, then $∀x(P(x)→Q)$ is semantically equivalent with $∃xP(x)→Q$. How to understand this statement. And also, need an example to show that the formula will not necessarily be equivalent if $x$ does occur free in $Q$.  I guess what struggles me is I'm not quite understand what does "free" mean in this case. But it would be better to have answer for the above question.
79811
In Wikipedia,  in the last axiom of the Natural deduction system, it says "From [accepting $p$ allows a proof of $q$], infer $(p \to q)$." Isn't that a tautology? In the big table "Basic and Derived Argument Forms" that follows, I see it using a new symbol "$\vdash$" to express "therefore". Isn't that something can be done by the symbol "$\to$"? For example, what's the difference between "$(p \land q) \vdash p$" and "$(p \land q) \to p$"?  An interesting symbolic representation of the last axiom of NDS is that $(p \vdash q) \vdash (\vdash p \to q)$. Which I don't understand why is made so complicated.    What also puzzles me is in the  that follows. Since you list "$A$" as a promise, which is basically saying "assuming $A=\text{True}$", then why you have to go around such a big circle, using disjunction, conjunction, and conjunction elimination to prove back that "$A$" is true? If it is me, I would say "assuming $A=\text{True}$, we will know that $A=\text{true}$. Bang! Now we get $A \vdash A$, so we have $\vdash A \to A$." What's wrong in this deduction?
67103
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?
15402
In my book about Logic, which is called 'Language, Proof and Logic', by the way, there is explained that the conditional $ P \rightarrow Q $ is equivalent with $\neg P \lor Q$.   that gives as answer: 'just compare the truth tables and you can see that they are equivalent'. However, I would like to know how to prove this using formal propositional logic. This should be possible, right?
17837
I studied propositional logic, and everyday I see applications of what I learned on the internet, in mathematical books and miscelaneous resources.        One particular case is sentences in the form $a \rightarrow b$ , that is, material implications. While I fully understand the mathematical logic approach to the study of those kinds of formulas, sometimes I get confused about what someone ( using implication sentences in the ordinary language ) is trying to say.        In ordinary language, I often see affirmations in the form of an implication.  One example would be someone affirming that "the moon is made of cheese" $\rightarrow 2+2=4$. [ Kleene, Mathematical Logic ].      I perfectly understand that if we choose the interpretation that gives the antecedent as false ( and the consequent as true ) we get that the whole implication is also true, but I would like to understand EXACTLY what he is trying to say when he AFFIRMS a sentence in the form of an implication. Is he saying that under the commonly agreed interpretation of that sentence ( antecedent = false, consequent = true ), that implication becomes true?      Another example, totally different, would be someone affirming that $x \in \mathbb{N} \rightarrow x \in \mathbb{N}$ or $x \in \mathbb{Z}_{&lt;0}$. Is he now affirming the same kind of sentence or saying something totally different?      When he affirms that sentence, is he saying that he finds the sentence to be a tautology?   Why didn't he say that?          Then, with these two examples we have an affirmation of the same kind of sentence (implication) meaning completely different things, in one case he affirms to say that under the commonly agreed interpretation of the sentence, it is true. In another case, he affirms to say that there's no interpretation that makes the sentence false.       Am I missing something? There is really that ambiguity in ordinary language, when people are affirming sentences?  Can't I just take some general instance of the affirmation of a sentence in the form $a \rightarrow b$ in ordinary language and know exactly what he's trying to say, regardless of the propositions he chooses the substitute $a$ and $b$ for?       Thanks a bunch.
118996
I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?  I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.  This  is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.   Thanks a lot in advance.
122936
I know I have to prove both sides here so: $\implies$: Suppose $|a| &lt; b$. Since $|a|$ is a positive number, we know $b$ is a positive number greater than $|a|$:  If $a$ is positive, $a &lt; b$ If $a$ is negative, $a &lt; b$  But for $|a| &lt; b$, it has to be that $a &gt; -b$. $a &lt; b$ and $a &gt; -b$, therefore $–b &lt; a &lt; b$ is true. $\impliedby$: Suppose $-b &lt; a &lt; b$.  If $a$ is positive, then $b$ is a greater positive number, and $-b$ would be less than $a$ If $a$ is negative, then $b$ can only be a greater positive number for $-b &lt; a$ to hold  So, $b$ is a greater positive integer than $a$. Therefore, it must be that $|a| &lt; b$. This was what I came up with but it seems convoluted and kind of forced with the cases. Any suggestions?
115892
I've been studying Discrete Mathematics from Kenneth Rosen's book. In that book, there is this "$(p \vee q) \land (\neg p \vee r)) \rightarrow (q \vee r)$" proposition if we can say so.   The book wants me to prove that this proposition is a tautology.  I've searched on the Internet and also on here but couldn't figure out how to prove this without a truth table. There is another asked question but that is so confusing and is not really illustrative.  Would you help me to prove this, please? Thank you in advance. :)
123456
I have the following question: "Explain the concept of the soundness of a rule of a proof system with respect to the truth tables"  Would it be correct to state the following:  "A rule of a proof system is sound with respect to the truth tables iff the conditions of the rules (requirements of the rule) ensure that the final formula will behave according to it's intended truth table."?  This is a question in a practice exam, and I can not find a specific mentioning so far, by scanning Google and Proof Wiki.  What would the correct answer be?  Thank you.