New answers tagged

1

What's meant is $$\exists N\in \mathbb{N}: \forall n\ge N:\frac{n-2}{n^2-4n+2}<0.07$$ so that the statement holds for all $n$, except at most finitely many (the ones that are $<N$ for which we have no info) The negation is $$\forall N\in \mathbb{N}: \exists n\ge N:\frac{n-2}{n^2-4n+2}\ge 0.07$$ which implies there are infinitely many $n$ for which the ...


1

If you're saying $[\forall y \in B (f(y) = x)] \implies [\forall y \in B (\exists y \in B (f(y) = x))]$, this is indeed true. If you're saying $[\forall y \in B (f(y) = x)] \implies [\exists y \in B (f(y) = x)]$, this is not necessarily true (for consider $B = \emptyset$).


2

Is it legitimate when I just write: $\forall a \in X:P(a)\rightarrow \forall x \in X: P(a)$ It would less confusing if you wrote equivalently: $\forall a \in X:P(a)\rightarrow \forall x \in X: P(b)$ This implication is not necessarily true. It would be true if $X=\{b \}$. It would be false if we had $X=\{c\}$, $c \neq b$ and $\neg P(b)$.


3

The meaning of $\forall$ is given in natural language. "$\forall$" and "$\text{for all/any/each}$" live in different realms. Natural language has meaning before we start constructing the formalism, and so it can be used to give the semantics of "$\forall$" without being circular or redundant. Your translated bit of text is ...


2

Should we instead write $$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:((a,b')\in R\implies b=b')).$$ Yes. You would use this to prove that the set of ordered pairs $R$ is a function mapping each element of $A$ to a unique element of $B$. If you have established or simply assumed that $R$ is a function mapping each element of $A$ to a ...


0

I'd say that your version is actually better. In the context of the definition of functions, where the author is establishing the uniqueness of $R(a)$ using Russell's definite descriptions, one should understand the last quantifier to range over the whole following conditional, but as written it could be interpreted as ranging over the antecedent only, and ...


1

It is in fact guaranteed to be $\Sigma^0_3$. Remember that each arithmetical class is closed under (finite) conjunctions and disjunctions, and here the left conjunct is $\Pi^0_1$ hence a fortiori $\Sigma^0_3$ while the right conjunct is $\Sigma^0_3$. To see this explicitly, note that the formula is equivalent to $$\exists a\forall b\forall x\exists c(P(x)\...


0

Let $V$ denote the partial function from the question. Suppose $n\notin A$. Then $\exists x_0 \forall y_2 \neg T(n, x_0, y_2)$. Then the domain of $V(n,-)$ is a subset of $\{x\in N: x < x_0 \}$ (hence finite). Here's why nothing outside this finite set can lie in the domain. If $x \ge x_0$, then the condition $$\forall y_1 < x \exists y_2 T(n,y_1,y_2)$$...


0

∀x∃y∃z (3x=150y-39z)≡T For all integer x, Let y=6x, z=23x LHS=3x RHS=150(6x)- 39(23x) ∴LHS=RHS,for ∀x∈Z


1

Universal elimination when $∀$ is not the main operator. You cannot apply UE when the quantifier is not the main operator. You have to derive $\forall x A(x)$ in order to use it to "detach" $B$ using ($\to$-E): $\forall x A(x) \to B$ --- premise $\lnot \exists x (A(x) \to B)$ --- assumed [a] $\lnot A(y)$ --- assumed [b] $A(y)$ --- assumed [c] ...


3

The quantifiers $\forall x\in D~$ and $\exists x$ each bind all free occurrences of $x$ inside their scope. However, any such occurrence can be bound to only one quantifier, and its immediate container has precedence. Since such occurrences of $x$ inside the scope of $\exists x$ are bound to that quantifier, therefore they are not considered free inside the ...


1

See Universal generalization. The rule formalize the intuitive sound argument that, if we can assert that a "generic" object $x$, i.e. an object whatever, has property $P$, then everything has that property. The "generic" restriction must be expressed with the restriction that nothing must be asserted in the "context" [the ...


1

The formula $$\forall a\forall b\Big(f(a)=f(b)\rightarrow a=b\Big)$$ states that the function $f$ is injective: meaning that, for any two elements of its domain, if $f(a)=f(b)$, then you actually have that the elements you started with are one and the same, $a=b$. But your formula is not actually wrong! It just has a redundant part, that is, $$\forall a \...


0

I don't know why you insist that $(n,x)$ not be defined when it doesn't take the value $1$. I prefer to think of it as a total function that takes the value $0$ when $\forall y_1 \le x, \exists y_2 T(n,y_1,y_2)$ is not true, i.e. when $\exists y_1 \le x, \forall y_2, \neg T(n,y_1,y_2)$. Then the question is whether the set of $x$ for which $(n,x)$ maps to $1$...


1

Russell's paradox relies on existential instantiation and universal instantiation, and these are not tools for the amateur mathematician. They are tools for everyone. Existential instantiation: If you are told that there exists an object with a certain property, you can give a unique name to one such object. Universal instantiation: If you are told that (1) ...


1

Suppose that $V$ is our universe, and $A$ is a subset of that universe. We can relativise the formulas to $A$ by recursion: If $\varphi(\vec x)$ is an atomic formula, then $\varphi^A(\vec x)$ is simply $\varphi(\vec x)\land\vec x\in A$. (Here $\vec x\in A$ is understood as the conjunction of $x_i\in A$ for the variables appearing in $\vec x$). If $\varphi=\...


0

The basic fact is to define what does it mean for a structure (or: interpretation) $\mathfrak A$ to satisfy formula $\varphi$ with variable assignment function $s$. In symbols: $\mathfrak A,s \vDash \varphi$ (alternatively: $\mathfrak A \vDash \varphi[s]$). Intuitively, the definition means that the "translation" of $\varphi$ determined by $\...


0

The definition is not that there is a different succession. The definition is that there is some succession. This may be different or identical to the original one.


2

To disprove the first one set y = x. To determine the second one, use the equation in P to reduce Q to a single variable and make your decision. Explain why the third one is nonsense.


-1

It's a good exercise to see the natural bijections worked out in some detail, but as I noted in your other reply (Categorical semantics explained – what is an interpretation?), padawan, there is an additional minor detail that you left out, which should be made clear. While the natural bijection for the $∧$ instantiates to bijections between the hom sets $𝐋(...


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