I am struggling to prove, without using truth tables, that the statement is a tautology.  [(p→q)∧(q→r)]→(p→r)                                   My work so far...  ¬[(¬p∨q)∧(¬q∨r)]∨(¬p∨r)                                                                                           ¬(¬p∨q)∨¬(¬q∨r)∨(¬p∨r)                                                                                      (p ∧ ¬q) ∨ (q ∧ ¬r) ∨ (¬p ∨ r)
I have been attempting to use identities to get to the answer but I am unable to get anywhere. Here is the equation I am trying to prove tautological without using truth tables:  $[(p\rightarrow q) \land (q \rightarrow r)] \rightarrow (p \rightarrow r )$  And here is my attempt at using identities to work my way up. (a link to a page with a table of logical equivalence identities: )  $\lnot [(p \rightarrow q) \land (q \rightarrow r)] \lor (p \rightarrow r)$  $\lnot [(\lnot p \lor q) \land (q \rightarrow r)] \lor (p \rightarrow r)$  $[\lnot (\lnot p \lor q) \land \lnot (q \rightarrow r)] \lor (p \rightarrow r)$  $[(p\land \lnot q) \land (q \land \lnot r)] \lor (p \rightarrow r)$  I don't know where to go from there.
Both options affect colors in a similar way, but I can't figure out the difference. How to make a WMS GetFeatureInfo request in Leaflet? I am trying to make a WMS GetFeatureInfo request using Leaflet. I can not find something in the documentation to describe this process. Can you please provide with an example demostraing how to do something like this? Where are the Unity desktop wallpapers located? The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? Can you please explain this to me?          Could you please explain this to me?   I am unable to figure out which to use which situation. I did google, and some posts say they are both the same, even if the second one is more formal. Other posts say could is the past tense of can, but in the above example, I don't think could is used as past tense of can.  What is correct way to use these words? The SPListItem being updated was not retrieved with all taxonomy fields On a whim the other day, I changed my meta name from Lord Torgamus to Popular Demand.  I just noticed that the old name has "stuck" to the author tag of ; everywhere else on meta, my new name displays as expected.  Is this a feature?  Is it a bug?  Mostly I'm just curious, as I can't think of a strong reason in favor of keeping this behavior or a strong reason for changing it.  A question somewhat like this has been raised before as , but the previous poster implied that the issue had something to do with migration, which my situation shows is not the case. Was there any reason why Dr Soong chose for Data's eyes to be yellow?  I can understand that he wasn't going for a completely human-looking android based on his skin color, but why yellow eyes? Do vulnerability and resistance cancel out? Equivalent definitions for simple connectivity Centering with \includegraphics not with \begin{figure}? I mess up with things when I try to use the figure-thing: the picture is not appearing with the correct alignment (apparently because I forget the float package). Now I want to use only the graphicx package with \includegraphics. How can I center the \includegraphics without figure?  Not working or doing centering to the document afterwards.  \includegraphics[width=0.5 \textwidth]{./Pictures/r.png}{\centering} \includegraphics[width=0.5 \textwidth]{./Pictures/r.png}\centering Center figure that is wider than \textwidth I have a figure that is wider than the \textwidth of my document. (I don't want to change the \textwidth.) As a result, the leftmost part of the figure is flush with the left margin, while the rightmost part spills into the right margin too much. I'd like to center the figure, so that it will exceed the \textwidth equally on both sides. When is the past perfect exactly needed? Export tikz as image file Does iOS have a firewall? Does iOS support a built-in firewall like macOS does? I don't see anything about it in Settings. I am going through Shankar's Principles of Quantum Mechanics and am having trouble finding the free particle propagator $U(t)$ that satisfies $$\lvert\psi (t)\rangle = U(t)\lvert\psi (0)\rangle$$ due to the degeneracy of the $E$ eigenkets. Shankar says that if the Hamiltonian has degenerate eigenvalues, we change the propagator equation from $$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$ to $$U(t) = \sum_{E} \sum_\alpha \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar} $$ where $\lvert E, \alpha \rangle $ are the orthonormal eigenkets for the $E$ eigenspace. This makes sense to me, because it seems like whenever we deal with degeneracy to use a projection. He also says for a Hamiltonian with no degeneracy to change the sum into an integral, so I imagine that for a Hamiltonian with a continuous, degenerate spectrum we would have this propagator: $$ U(t) = \sum_\alpha\int_\Re dE\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$  I tried applying this to the free particle and had some trouble. In the book, after solving for the eigenkets and eigenvalues, $\lvert E\rangle = \lvert p\rangle$ and $p = \pm\sqrt{2mE}$, he chooses the propagator as a function of the momentum eigenvalues $p$ so that $$U(t) = \int_\Re dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m).\tag{5.1.9}$$ I understand how he got to this, but I'm having some trouble because this doesn't appear to be the same expression that I have above. Splitting the integral up, we have $$U(t) = \int_{-\infty}^0dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ The left integral can changed to an integral over $-p$, and the bounds switch from $0$ to $\infty$ such that: $$U(t) =  \int_{0}^\infty d(-p)\lvert -p\rangle\langle -p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ Now we can substitute and interchange $\lvert p\rangle = \lvert E, + \rangle$,  $\lvert -p\rangle = \lvert E, - \rangle$, $E = p^2/2m$, and $dp = \pm mdE/\sqrt{2mE}$ to get: $$U(t) = \sum_{\alpha = \pm} \int_0^\infty dEm/\sqrt{2mE}\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}. $$ This is the result that we are asked to prove in Exercise 5.1.1. Why is this not the same as the first expression for the propagator? Where did the extra $ m/\sqrt{2mE} $ come from? Am I just wrong in assuming that that is the general expression for a Hamiltonian with continuous and degenerate eigenvalues? Escaping both forward slash and back slash with sed What prevents this magnetic perpetuum mobile from working? Please stop using SoftwareEngineering.SE as your toilet bowl Stack Overflow users have been using the "off topic - belongs on SoftwareEngineering.SE" close reason as an alternative to all the other close reasons. It'd be helpful if people actually read the FAQ and the six  before voting to migrate questions like:   , no redeeming value to favorite lists that become argumentative , "help my exact situation" questions are too localized , fails 5 out of the 6 guidelines , fails 4 out of 6 , too localized and too vague , too broad   Absent a reversal of the Stack Overflow community from viewing SoftwareEngineering.SE as the Stack Overflow cesspool, can the "off topic - belongs on SoftwareEngineering.SE" close reason be removed in favor of manual migration by moderators who can actually spend the 10 seconds to determine if the question should be closed outright first?    To say that SoftwareEngineering.SE is such a nebulous place and nobody knows what its scope is as the reason why crappy questions keep getting migrated there is to miss the point. Good questions are good questions on all of the sites, and questions that are too vague, that are too argumentative, too localized, and too broad are universal close reasons. There is no special dispensation for SoftwareEngineering.SE to have bad questions, even if you personally think the questions on there suck.  Even the six subjective guidelines for subjective questions are universal guidelines: they affect all sites. So even if you had no idea what the consequences of using the "off topic - belongs on SoftwareEngineering.SE" close reason were, there are still all of the other close reasons for which bad questions, like the ones above, can be closed. How do I create a very long delay with Redstone?