A similar question had already been asked, but the solution involves steps I am unfamiliar with. in class, we have only been exposed to intro and elim rules, as well as contradiction rules.  Here is the original question:  I am not familiar with "DS" as a method
123648
It's all in the question really. I am working on a proof in Fitch for a class, but I am very much stuck.  I am proving the tautology that "(P → Q) ↔ (¬P ∨ Q)", and I have already finished half of it, but now I must prove that "(¬P ∨ Q)" implies "(P → Q)". I can't seem to get anywhere.   I try to set up a proof by cases where I assume in different subproofs "¬P" and (in the other) "Q", but then I must prove "P → Q" from those. It seems even more difficult. Any help would be appreciated.
47039
Just doing this for revision, seems much harder than it should be, should I use $x=\tan u$ ? Any help appreciated.
21353
Using only rules of natural deduction, I am trying to prove that $$p \implies q \vdash \lnot p \lor q$$ but am having a lot of difficulty. I was able to prove the other direction.   Could anyone show me or point me to the right direction?
121569
From this paper:     Shouldn't $du$ be $dt$? And also how do you get from that line to the final result if $du$ is not $dt$?
18184
I sense that the following simple argument is invalid, but cannot figure out why. Base Case:No variable or constant is an initial segment of a term, since the only terms that begin with variables or constants have length 1, and so lack initial segments. Inductive Case:Suppose s is an initial segment of a term, t, where s is $f t_1 t_2 \cdots t_n$ for terms $t_i$. Then t is su, i.e., the string, s, followed by the string, u, for some non-empty string u.  But, then, t is not a term, since an n-ary function symbol followed by n terms, followed by a non-empty string is a not a term. Hence, no term is an initial segment of any other. I did not make use of the inductive hypothesis, that no ti in s is the initial segment of any term.  This is what makes me think I cheated somehow.  Assuming that I did, I wonder if someone could show me how to prove the conclusion by means of induction on the complexity of terms without assuming that the tis in t are not initial segments of any terms (i.e., by merely assuming that the tis in s are not). Any feedback would be greatly appreciated!  Thanks in advance.
123933
I am not a mathematician, I'm a linguistics PhD student. As part of my research I need to put various convoluted sentences through various syntactic transformations and see then check whether people think they are true or not. Mathematical statements (well, some of them) suit my purposes very well, because they are less context dependent and can be straightforwardly assigned a truth value (i.e. be deemed true or false). The problem is that I'm not a mathematician. When these sentences get a bit convoluted, I have a bit of a problem knowing whether they are true or false myself (before they undergo various syntactic transformations).   I have a particular sentence which states that if a given number is:   an integer divisible by 7 (meaning it will yield an integer if divided by 7) a square number   then it is divisible by 49. I intuitively believe this to be correct (although I can't explain why). Is this actually true? I don't want to waste everybody's time by starting with an untrue untransformed sentence.
14375
starts with      I know that this isn't the first post with this problem, but I am   stuck with this for a day (total, not 1 day in a row). Somehow I can't   get this to work, maybe it is just 1 little adjustment that I need to   make, but I can't seem to find it :(   I'm sure the question , both for immediate answering and for people later having the same kind of problem.  But I'm afraid removing it might be considered rude by the asker as he might see the question with the introduction both more humble and more accorded to his personal writing style (especially in a case where the code follows immediately after this introduction).  Should I proceed and remove this lengthy useless introduction or just not bother ?
19748
I am trying to solve this problem and looking for any tips. I have 5 D's, 6 E's and 3 F's and I have two conditions.  The first D must be before the first E.  The first E must be before the first F.  I thought I could solve this through determining the ways to make a string of letters with a D before an E before an F, however, that is not enough because it does not cover instances like DEFDEF... or DEFFD... since there can be letters that come after the initial string that meets the conditions.  I also contemplated using the Inclusion Exclusion principle as follows  cases where a D comes before an E + cases where an E comes before an F subtract their intersection (which I'm not sure how I'd calculate) however I still cannot account for cases where there are letters after the initial D, E, F...  Any ideas?
11488
so I used the root test, but i'm not quite sure if i'm allowed to. I think im performing the operations correctly,a and i keep ending up with $(1)^{\infty}$. So really my question is am i performing the operations wrong or do i have to use a different test?
14868
I talked to my professor and they said that it can be disproven using an example, specifically using a picture to explain why it statement is false. Can anyone help me out?
86450
$(s \to p) \lor (t \to q) \vdash (s \to q) \lor (t \to p)$ This has been giving me a terrible headache. I have no idea how to go about proving this. I'm not asking for a complete solution, I just need some pointers to prove this with natural deduction...
11364
In exam class, how to print Answer instead of Solution when using  \begin{solution}  \end{solution}   in question-answer format?
87037
My intention is to create equation which looks similar to one of examples below (I prefer the second one).     There should be an arrow from JKL to DEF and GHI and from GHI to DEF.   Neither alignment nor equation environment have to be used. Anything that will do is appreciated. Cheers.
122666
I'm taking a course from Stanford in Logic. I'm stuck with an exercise where I'm doing some proof. The Fitch system I'm given only allows   $ \land $ introduction and elimination $ \lor $ introduction and elimination $ \to $ introduction and elimination $ \equiv $ introduction and elimination $ \lnot$ introduction and  $ \lnot \lnot $ elimination   but not    $ \bot $ introduction and elimination   I've been struggling to prove the law of excluded middle (``p ∨ ¬p`) within this system. All of the proofs I've seen online make use of $\bot$ elimination to prove it by contradiction:  | ~(p | ~p)  (assumption) | | p        (assumption) | | p | ~p   (or introduction) | | $\bot$ | | | ~p       (assumption) | | p | ~p   (or introduction) | | $\bot$ | | (p | ~p)   (proof by contradiction)   Since I can't use $\bot$ elimination, I was trying to do this with $ \lnot$ introduction:  | ~(p | ~p)  (assumption) | | ??? | ~(p | ~p) =&gt; x | | | ??? | ~(p | ~p) =&gt; ~x | p | ~p     (not introduction)   but I just can't get it. I can't figure out how to get x and ~x.  Is this provable without $\bot$ elimination? If so, what are the steps?    Background:  What I'm ultimately trying to prove is : Given p ⇒ q, use the Fitch System to prove ¬p ∨ q. I'm able to prove that p ⇒ ¬p ∨ q and that ¬p ⇒ ¬p ∨ q, but I need to be able to plug p ∨ ¬p into the proof to be able to do 'or elimination'.
101413
Here's another of Tomassi's exercises I can't solve (Logic, page 106):  ~ (~P &amp; ~Q) : P ∨ Q   I have to use natural deduction and the only rules I know are:   assumptions, modus ponendo ponens, modus tollendo tollens, double negation, reductio ad absurdum, conditional proof, v-introduction, v-elimination, and introduction, and elimination.   Tomassi's proof consists of 14 steps.  Thanks for the help.
97484
Would anyone mind explaining me how proof by contradiction works? I have a very vague understanding of it so I always avoid using it when it comes to discrete math. From what I've seen its something that is extremely useful when it comes to proofs.   From what I understand proof by contradiction is the following. Given an example I have to prove it to be true. I take the complete opposite from it and prove it to be false(?), from then on I can assume that the original example is true(?). If anyone would mind correcting my understanding, please do. Also, what would be a good step-by-step guide I can have in my mind when using it?
93154
I'm having problems with this; the strategy is to show the left side is in the right side, and vice versa. Thus far I can only show that both sides of the equality are in A\B. Can I use "the empty set = A and not A" as a fact? Because then it works. Or just show me how to do it!