Let $x$ be an individual variable. What is $x = x$?  An open sentence? Or a true proposition? Well, I think it's an open sentence, due to the presence of the free variable $x$. I do not know if I'm wrong. I ask for help.
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I have a doubt. I need help.  Can the basic axioms of equality be presented as "open axioms"?  I) (reflexivity) $\qquad x = x$              II)  (Substitutivity) $\qquad (x = y) \to \big(F (x, x) \to F (x, y)\big)$
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I found this proposition and don't see exactly as to why it is true and even more so, why the converse is false:  Proposition 1. The equivalence between the proposition $z \in D$ and the proposition $(\exists x \in D)x = z$ is provable from the definitory equations of the existential quantifier and of the equality relation. If $D = \{t_{1},t_{2},...t_{n}\}$, the sequent $z = t_{1} \vee z = t_{2} \vee z \vee ... z = t_{n}  \vdash z \in D$ is provable from the definition of the additive disjunction $\vee$.  The converse sequent $z \in D \vdash z = t_{1} \vee z = t_{2} \vee ... z = t_{n}$ is not provable.  The author goes on to say:  "We adopt the intuitionistic interpretation of disjunction. With respect to it, one can characterize a particular class of finite sets"  On the first part of the proposition, well I am not sure what the point is as if we take an element z in D, then we could just call this element x and hence this x = z.  Is there something more to this?  On the second part of Proposition 1 since $z = \text{ some } t \in D$ since t is in the set of axioms and $t = z$, then $t \vdash z$ as z is derivable from t.  For the converse if $z \in D$ then why wouldn't z = some $t \in D$?  Is this because of the Incompleteness theorem?  That perhaps there D as a set of axioms has some consequence which can not be proven by the set of axioms in D?  Or perhaps I am way off here.  Any ideas?  Thanks,  Brian
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Sometime like this case I don't know if I can put the name before or after the label (movie in this sentence)...  The movie &quot;Die Hard&quot; or The &quot;Die Hard&quot; Movie? The variable &quot;X&quot; or The &quot;X&quot; variable?  Are they both correct? What is their difference? What is the application of each? My question is not about &quot;the&quot; (), but the order of the &quot;movie&quot; and &quot;Die Hard&quot;
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How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$?  I have already recognized one problem caused from this: it  follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?  Best regards and thanks, John
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Really stuck with this, \begin{cases}  2tx'(t)-1=x^2t \\  x(t=1)=1 \\  \end{cases} I don't understand the variables. What does it mean that $x$ differentiated have $t$ as a function value, like this: $x'(t)$? Don't get it :(
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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 have encountered comparisions of variables to string literals multiple times over the years which had one character prefixing the variable and the literal, e.g.  if [ "x$A" = "xtrue" ]; then   in order to check whether $A is "true".  I assume this is done to achieve shell compatibility or to work around a longterm bug, an unintuitive behavior, etc. Nothing obvious comes to mind.  Today I figured I want to know the reason, but my research didn't turn up anything. Or maybe it's just me making something out of a rather frequent exposure to rare occurances.  Is this practice still useful, maybe even best?
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I do not understand the relation of a normal variable "x", which is to me just a placeholder for an element of a set, and a random variable, which is a mapping from the set of all possible events to $\mathbb{R}$.  To make the question more concrete, some parts I am struggling with:   Normal functions with variables can be evaluated, e.g. $f(x)=2x$ will plot to a line, can I do this with a function of a random variable? Furthermore the randomness part seems to be missing, why is it not neccessary to define how the random variable obtains its random value?     Edit: now that I feel I understood the definition, I cannot say what exacly what was missing for understanding. Nevertheless reading the following links, the comments and answer here, did the trick.
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What is ${\partial\over \partial x_i}(x_i !)$ where $x_i$ is a discrete variable?  Do you consider $(x_i!)=(x_i)(x_i-1)...1$ and do product rule on each term, or something else? Thanks.
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When is it appropriate to use an unsigned variable over a signed one? What about in a for loop?  I hear a lot of opinions about this and I wanted to see if there was anything resembling a consensus.   for (unsigned int i = 0; i &lt; someThing.length(); i++) {       SomeThing var = someThing.at(i);       // You get the idea.   }   I know Java doesn't have unsigned values, and that must have been a concious decision on ' part.
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I can't understand difference between $\varnothing$ and $\{\varnothing\}$. What is difference between $x$ and set that has only $x$ as element? I think the concept of set is meaningful when it has more than 1 element. If "$\varnothing$ and $\{\varnothing \}$ is different" is right, then I have another question. When we see the set $\{\varnothing,\{\varnothing\}\}, \varnothing$ and $\{\varnothing\}$ exist simultaneously. It means two empty sets exist. One empty set exists as $\varnothing$ itself and one empty set exists as element of $\{\varnothing\}$. I can't understand this situation. What am i misunderstanding?
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My question is told in a few words: Why do you write $E[X]$ in square brackets instead of something like $E(X)$? Probably it is not a "function". How would you call it then? This question also applies for $Var[X]$.
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I can write  VAR=$VAR1 VAR=${VAR1} VAR="$VAR1" VAR="${VAR1}"   the end result to me all seems about the same. Why should I write one or the other? are any of these not portable/POSIX?
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Taken from :  Show that it cannot be the case that $U=X+Y$ where $U$ is uniformly distributed on [0,1] and $X$ and $Y$ are independent and identically distributed. You should not assume that X and Y are continuous variables.  A simple proof by contradiction suffices for the case where $X$, $Y$ are assumed discrete by arguing that it always possible to find a $u$ and $u'$ such that $P(U\leq u+u') \geq P(U\leq u)$ while $P(X+Y \leq u) = P(X+Y \leq u+u')$.  However this proof does not extend to $X,Y$ being absolutely continuous or singular continuous. Hints/Comments/Critique?
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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.
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Question    Suppose we are dealing with real valued functions $f(x)$ of one real variable $x$. We say that the limit of $f(x)$ at point $a$ exists and equals to $L$ if and only if  $$\exists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)$$  and show this by the symbolism  $$\mathop {\lim f(x)}\limits_{x \to a}  = L$$  Now, what do we say when we want to state that the limit of function $f(x)$ does not exist at $a$? In fact, what is the negation of the above statement? I am interested to obtain the negation with a step by step approach using tautologies in logic. To be specific, I want to start from  $$\neg \left[ {\exists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)} \right]$$  and then go through it to get the final form of negation (See the example below).    My Thought  I just wrote down the two following negations without going through a step by step approach.  $$\forall L,\exists \varepsilon  &gt; 0:\left( {\forall \delta  &gt; 0,\exists x:0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| \ge \varepsilon } \right)$$  or  $$\nexists L:\left( {\forall \varepsilon  &gt; 0,\exists \delta  &gt; 0:\left( {\forall x,0 &lt; \left| {x - a} \right| &lt; \delta  \implies \left| {f(x) - L} \right| &lt; \varepsilon } \right)} \right)$$  I want to know weather these are true or not.       Example    I will give an example of what I mean by a step by step approach. Consider the following statement  $${P}\implies R$$  I want to take a step by step approach to obtain the negation of the above statement. Here is what they usually do in logic  \begin{align} \, \neg \left( {P \implies R} \right) &amp;\iff \neg \left( {\neg P \vee R} \right) &amp; \text{Conditional  Disjunction} \\ \qquad \qquad \quad &amp;\iff \neg \neg P \wedge \neg R &amp; \text{Demorgan's  Law} \\ \qquad \qquad \quad &amp;\iff P \wedge \neg R &amp; \text{Double  Negation} \end{align}
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Can anybody give a clear explanation of how variable assignment really works in Makefiles.   What is the difference between :   VARIABLE = value  VARIABLE ?= value  VARIABLE := value  VARIABLE += value   I have read the  in GNU Make's manual, but it still doesn't make sense to me.