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## Hot answers tagged quantifiers

5

In order to evaluate the truth value of $∃x \ ∀y \ ∀z \ P(x,y,z)$, it is useful to "read" it : "there is a positive real $x$ such that, for every (positive reals) $y$ and $z$ it is true that $xyz=1$. The reasoning is : assume that $x > 0$ exists such that .... From $xyz=1$ we get $yz= \dfrac 1 x$ (we can do it because we have $x > 0$) for every $y,... 4 Ok, several ways to do this: Probably the most straightforward way is to say that there are three distinct orange apples and no more:$\exists x \exists y \exists z (A(x) \land O(x) \land A(y) \land O(y) \land A(z) \land O(z) \land x \neq y \land x \neq z \land y \neq z\land \neg \exists w (A(w) \land O(w) \land w \neq x \land w \neq y \land w \neq z))$... 3 would the first proposition indicate that there is an x that is both Blue(x) and a Circle"(x) or could the x of Blue(x) be distinct of the x that satisfies Circle(x)?" The first proposition says that there is an$x$such that both$Blue(x)$and$Circle (x)$holds simultaneously. The second proposition says that there is an$x$such that$Bed(x)$holds AND ... 3 I think of interpretation as a two-stage process. First, translate the symbols into "mathematical language" without referencing the quantified terms, and then coax it into natural language. For instance, $$\forall x\ E(T,x)$$ is "for everything, Tom eats it", which I revise to "Tom eats everything." But if it's something where$x$is referenced in both ... 3 The first statement says: for each$j$in$\{1,2,3\}$, you can find a real number$a$(which number may depend on which$j$) with the property that the$j$th function maps$a$to$1$. Presumably, you have three functions called$f_1$,$f_2$, and$f_3$, and the first statement tells you that each and every one of these three functions "hits" the value$1$... 2 The first one says that for every$j$in the set$\{1,2,3\}$, there exists an$a$in the set of real numbers dependent on$j$such that$f_j(a)=1$. The second, that there exists an$a$in the set of real numbers such that for every$j$in the$\{1,2,3\}$, no matter which one, we have$f_j(a)=1$. Can you take it from here? Hint: 2 I assume you meant$\forall x(P(x) \land S(x))$(with$S(x)$inside the scope of the quantifier$\forall x$), not$\forall x(P(x)) \land S(x)$. If you say$\forall x(P(x) \land S(x))$, this means "For all entities in the domain,$P$holds and$S$holds". This statement is false in the domain$\{1,2,3\}$, since neither$P$nor$S$hold of$3$. If you want ... 2 Statement 1 is true and statement 2 is false, because the order of quantifiers matter. As the existence quantifier in statement 1 comes after the "for all" quantifers, it claims existence after assignments have been made to the variables named in those. Contrary to that the existence quantifer in statement 2 comes first, and therefore claims existence of ... 1 1) There are at least two people who everyone knows. Domain = {People} My take in this.... for the 1) part is it valid to do something like this ∃𝒙∃𝒚∀z𝑷(𝒙, 𝒚,z). Where in my words i could be totally wrong.. there exist a pair (x,y) who everyone (z) knows. You must say: "There are some$x$and some$y$who are not the same people and every$z$... 1 One could use this tree proof generator to determine if these formulas are true or false. If a tree is generated then the formula is true. If a countermodel is generated then it is false. These are simple enough that they will reach a conclusion quickly. Here is the first one with a countermodel showing that it is false: https://www.umsu.de/trees/#%E2%88%... 1 Your best bet is to come up with interpretations of$P$that will let you think about whether the implications are true. For instance, in the first problem, you could let$P(x,y)$be the statement$|x|+|y|=0$where$x$and$y$are real numbers. Then it is true that$\exists x\exists y P(x,y)$. Is it also true that$\forall x \exists y P(x,y)$? Then try ... 1 So you have three functions, named by the indices$\{1,2,3\}$, which accept real values as arguments (aka. their domain is points on the real number line).$(\forall j \in \{1,2,3\}) ( \exists a \in \Bbb R)~~f_j(a)=1\$ Each index has at least one argument which makes its function equal one. These arguments need not be the same. Each function will ...

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