I don't get the idea behind it. There are some things in the truth table which disturbs me. As if $A$ is false, the $A\Rightarrow B$ is always true. I read that implication was the negation of ($A \land (\neg B) $). Thus indicating that if $A$ is true and $B$ false, then the implication is false and that is all we care about, the other cases are just a consequence of the definition. Is this right?
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.
90608
I am self-studying out of Velleman's "How to Prove It", and am trying to solve exercise 7, part a from section 1.5. The problem is to show that  $$ (P \to Q) \land (Q \to R) = (P \to R) \land ((P \leftrightarrow Q) \lor (R \leftrightarrow Q)) $$  By converting the right-hand side to the equivalent AND/OR statements I was able to simplify the expression to  $$(\lnot P \lor Q) \land (\lnot Q \lor R) \land (\lnot P \lor R)$$  which is equivalent to the left-hand side (confirmed through truth tables, Venn diagrams, and Wolfram Alpha), but I am having trouble showing it analytically. Clearly, the first part of this statement is identical to the left-hand side. But I cannot figure out the analytical steps to show that $\lnot P \lor R$ cancels out of the statement. I've tried distributing the statements into each other completely, but I keeping ending up back at the above expression.  By converting back to conditional statements, the fact that $\lnot P \lor R$ is dispensable becomes intuitive to me. That is, it makes sense that $P \to Q$ and $Q \to R$ already implies that $P \to R$. However, I want to be able to show it analytically based on the Boolean laws and I've hit a wall there. I'm sure I'm missing something simple, but it's driving me crazy. Any hints?
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.
28448
Here is my proof, I would appreciate it if someone could critique it for me:  To prove this statement true, we must proof that the two conditional statements ("If $\mathcal{P}(A)⊆ \mathcal{P}(B)$, then $A⊆B$," and, If $A⊆B$, then $\mathcal{P}(A)⊆ \mathcal{P}(B)$) are true.    Contrapositive of the first statement: If $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$  If $A \nsubseteq B$, then there must be some element in $A$, call it $x$, that is not in $B$: $x \in A$, and $x \notin B$. Since $x \in A$, then $\{x\} \in \mathcal{P}(A)$; moreover, since $x \notin B$, then $\{x\} \notin \mathcal{P}(B)$, which proves that, if $A \nsubseteq B$, then $\mathcal{P}(A) \nsubseteq \mathcal{P}(B)$. By proving the contrapositive true, the original proposition must be true.    To prove the second statement true, I would implement nearly the same argument, so that isn't necessary to write. So, does this proof seem correct? Also, was the contrapositive necessary? Or is there another way to prove the initial statement?
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 $(F \implies q) \equiv T$?
79283
Prove or give a counterexample:  $$(A ∪ B = A ∪ C) \land (A ∩ B = A ∩ C) \implies B = C.$$  I think this is true, but I am not sure how to show it. I don't know if there are any manipulations with unions and intersections that allows me to move things around. Thanks in advance for any hints.
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?
67531
The question gives the following: ¬(p ∨ ¬q) → (¬r ∧ (p → r))  With my own working, I have come to the answer of (p ∨ ¬q) ∧ (¬r ∧ ¬p) ∧ (¬r ∧ r). Can anyone tell me if this is correct?  Also, please tell me what Disjunctive Normal Form really means (without complicated terminology). The answer I got was from trying to break down what other similar questions have done. I still don't know what it really means.
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.
16168
Someone recently asked me why a negative $\times$ a negative is positive, and why a negative $\times$ a positive is negative, etc.  I went ahead and gave them a proof by contradiction like so:  Assume $(-x) \cdot (-y) = -xy$  Then divide both sides by $(-x)$ and you get $(-y) = y$  Since we have a contradiction, then our first assumption must be incorrect.  I'm guessing I did something wrong here. Since the conclusion of $(-x) \cdot (-y) = (xy)$ is hard to derive from what I wrote.  Is there a better way to explain this?  Is my proof incorrect?  Also, what would be an intuitive way to explain the negation concept, if there is one?
102950
I am reading introductory texts on logic and am having a hard time understanding intuitively logical implication.  Specifically, I am wondering if logical implication can always be determined by two statements, P and Q alone? I'll try to give an example to clarify my question.  For example if the statement P is $x = 2$, and the statement Q is $x^2 &lt; 6$, then determining the truth of P $\Rightarrow$ Q is fairly straightforward:   If $x$ does equal 2, then $4&lt;6$ and Q is true, so $P \Rightarrow Q$ is true. And if $x$ does not equal 2, $P \Rightarrow Q$ is vacuously true.  In this example, with two statements P and Q, I can evaluate the truth/falsity of the statement P implies Q. However, what if the statement Q is changed such that it is now the statement $y = 5$?  So now I am trying to prove from that $$x = 2 \Rightarrow y = 5$$  Here is where things get murky for me. Say $x$ is not equal to 2, then $P \Rightarrow Q$ is vacuously true. But what about when $x$ does equal 2, now the statement Q could be true, but it could also be false, depending on some 'context'.   For example, if I said at the start: "Here are two statements, P and Q. Know that x = 2 and y = 4." Given this 'context', $P \Rightarrow Q$ is false. But if I say "Here are two statements, P and Q. Know that x = 2 and y = 5", then now $P \Rightarrow Q$ is true.  My discomfort here is that I assumed with two statements P and Q, I can determine the truth/falsity of $P \Rightarrow Q$ with just this information. Is this the wrong assumption?  As a secondary question, how can the truthfulness of the statement $P \Rightarrow Q$ be variable, depending on context? I have read on some other questions that implication can be interpreted as a promise. So how could it be that depending on some context the promise is sometimes true but sometimes false? Wouldn't this actually mean the promise is not always consistent and is therefore always false?  Some further information in case it is relevant, I recognise that in the first example, the statement Q is 'dependent' in some way on the statement P, since the object under examination in both statements is $x$, whereas in the second example there are two different objects which I am trying to make a logical connection between, which I am thinking may not be valid for some reason. I also recognise that there is still 'context' in the first example, in that we need information on whether $x$ is equal to 2 or not, however in that case it doesn't matter, since $P \Rightarrow Q$ is always true anyway.  Thanks very much in advance.
68608
I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.  For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$  implies $B = C$ is true.   I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:    I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof.  On the other hand, I could be completely wrong about my reasoning.    Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.
93267
Implication does not appear to be associative:  a  b  c | (a -&gt; b) -&gt; c | a -&gt; (b -&gt; c) F  T  F | F             | T F  F  F | F             | T   Is $a \implies b \implies c$ evaluated as $a \implies (b \implies c)$ or $(a \implies b) \implies c$?
22049
I think that the question is clear from the title:     Are there sets $A$ and $B$ such that $A \in B$ and $B \in A$?   My feeling is that the definition of any two such sets will be circular (we can't define $B$ till we define $A$ and we can't define $A$ till we define $B$), and hence will be excluded by some axiom.  On the other hand, I can't see an inherent contradiction, though I wouldn't be surprised if some form of Russell's paradox pops up, pushed behind one more set of braces.
116659
In a  I am reading, the section on Propositional Logic says that a proposition is a statement that is either true or false, but not both true and false. Also, from  lecture online, the instructor says that we must be able to associate a truth value to a proposition.    The text I mentioned contains as an example of an assertion that is not a proposition the following:    (1) "this statement is false."    In the margin, the text says that the form of this statement makes it impossible to designate a truth value to it and the instructor in the lecture says simply that, "if [the statement] is true, then it is false, and if it is false, then it is true."    However, why exactly is it impossible to for (1) to have a truth value? What does it mean to say that if (1) is true, it is false, and conversely?    Response to Asaf Karagila As has been pointed out, I have already asked this question very recently  but it has not received proper attention. This question is one that I feel can be put to rest if only someone would provide an explanation that is direct and suitable for my level, which is that of a novice.
41729
"This sentence is false". Is this sentence true or false?   My attempt:  If this sentence were true, then what it says would be the truth , it implies that it is false which is a contradiction. if it were false then it is false that this sentence is false, hence it  is true which is also a contradiction. So this sentence can neither be true be true nor false. Is my reasoning correct?  PS: I would like to see more of such statements.
18157
I've done a truth table after reducing it to this and it seems to be equal to $\neg Q$:  $$\lnot Q \lor (\lnot Q \land R) = \lnot Q$$  But when I try to show it without a truth table (with just transformations), I end up in a loop between that and:  $$\lnot Q \land (\lnot Q \lor R)$$  Is there a way to show this is true without using a truth table? What am I missing?!  Thanks in advance!