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If $f$ is a continuous injective map from a compact space $S$ to a Hausdorff space $T$, then $S$ and $T$ are homeomorphic.
If two topological spaces are homeomorphic, then they are compact if and only if they are compact.
If $a < b$, then $a$ is a limit point of the open interval $(a, b)$.
If $a < b$, then $b$ is a limit point of the open interval $(a,b)$.
If $a < b$, then the closure of the open interval $(a, b)$ is the closed interval $[a, b]$.
The closure of the set of all real numbers greater than $a$ is the set of all real numbers greater than or equal to $a$.
The closure of the set of all real numbers less than $b$ is the set of all real numbers less than or equal to $b$.
If $a < b$, then the closure of the open interval $(a, b)$ is the closed interval $[a, b]$.
If $a < b$, then the closure of the interval $(a, b)$ is the interval $[a, b]$.
The zero polynomial is not irreducible.
If $p$ is an irreducible polynomial, then $p$ does not divide $1$.
The polynomial $1$ is not irreducible.
If $p$ is a nonzero polynomial that is not a unit, and if $p$ cannot be factored into two nonunits, then $p$ is irreducible.
If $p$ is irreducible and $p = ab$, then either $a$ divides $1$ or $b$ divides $1$.
If $b$ is irreducible and $a$ divides $b$ and $a$ is not a unit, then $a$ is irreducible.
The element $0$ is not a prime element.
A prime element is not a unit.
If $p$ is a nonzero element of a commutative ring such that $p$ does not divide $1$, and if $p$ divides the product of any two elements, then $p$ divides one of the elements, then $p$ is a prime element.
If $p$ is a prime element and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.
If $p$ is a prime element, then $p$ divides $ab$ if and only if $p$ divides $a$ or $p$ divides $b$.
The number 1 is not a prime element.
If $p$ is a prime element, then $p \neq 0$.
If $p$ is a prime element and $p$ divides $x^n$, then $p$ divides $x$.
If $p$ is a prime element and $n > 0$, then $p$ divides $x^n$ if and only if $p$ divides $x$.
If $x$ is a prime element, then $x \neq 0$.
If $x$ is a prime element, then $x \neq 1$.
If two elements of a normalization semidom are equal after normalization, then they are irreducible if and only if the other is irreducible.
If two elements of a normalization semidomain are equal, then they are associates.
If $a$ and $b$ are associated elements in a normalization semidom, then there exists a unit $u$ such that $b = ua$.
If $x$ is a normal element of a normalization semidomain, then $x^n$ is also a normal element.
If $p$ is a prime element, then $p$ is irreducible.
If $x$ is a unit and $p$ is irreducible, then $p$ does not divide $x$.
If $x$ is a unit and $p$ is a prime element, then $p$ does not divide $x$.
If $p$ is a prime element and $q$ divides $p$, then $q$ is a prime element.
If $a$ is irreducible and $b$ divides $a$, then either $a$ divides $b$ or $b$ is a unit.
If $a$ is not a unit and if every divisor of $a$ is either a unit or a multiple of $a$, then $a$ is irreducible.
An element $x$ of a commutative ring is irreducible if and only if $x$ is not zero, $x$ is not a unit, and for every $b$ that divides $x$, either $x$ divides $b$ or $b$ is a unit.
If $a$ and $b$ are non-units and $ab$ is a prime element, then $a$ or $b$ is a unit.
If $p$ is a prime element and $a$ divides $p$ and $a$ is not a unit, then $p$ divides $a$.
If $p$ is a prime element and $p$ divides the product of the elements of a multiset $A$, then there exists an element $a \in A$ such that $p$ divides $a$.
The modulus of a complex number is less than or equal to the modulus of the sum of that number and another complex number plus the modulus of the second complex number.
For any polynomial $p$ and any real number $r$, there exists a real number $m$ such that $|p(z)| \leq m$ for all complex numbers $z$ with $|z| \leq r$.
The offset polynomial of $0$ is $0$.
The offset polynomial of a polynomial with a leading coefficient is the product of the offset polynomial of the rest of the polynomial with the offset, plus the offset polynomial of the rest of the polynomial.
The offset of a constant polynomial is itself.
The polynomial obtained by shifting the argument of a polynomial $p$ by $h$ is the same as the polynomial obtained by shifting the coefficients of $p$ by $h$.
If $p$ is a polynomial and $p(x) = c p(x - a)$ for some $c$ and $a$, then $p$ is the zero polynomial.
The offset polynomial of a polynomial $p$ is zero if and only if $p$ is zero.
The degree of a polynomial $p$ is the same as the degree of the polynomial $p(x + h)$.
The polynomial $p$ is zero if and only if its size is zero.
For any polynomial $p$, there exists a polynomial $q$ such that $p(x) = q(x + a)$ for all $x$.
If there exists an $x$ such that $P(x)$ is true, and there exists a $z$ such that $P(x)$ implies $x < z$, then there exists a real number $s$ such that $y < s$ if and only if there exists an $x$ such that $P(x)$ is true and $y < x$.
If $z$ is unimodular, then $z + 1$, $z - 1$, $z + i$, or $z - i$ has modulus less than 1.
If $b \neq 0$ and $n \neq 0$, then there exists a complex number $z$ such that $|1 + bz^n| < 1$.
The distance between two complex numbers is bounded by the sum of the distances between their real and imaginary parts.
If the sequence $(s_n)$ is bounded, then it has a convergent subsequence.
If $p$ is a polynomial and $z$ is a complex number, then there exists a $\delta > 0$ such that for all $w$ with $|w - z| < \delta$, we have $|p(w) - p(z)| < \epsilon$.
There exists a complex number $z$ such that for all $w$ with $|w| \leq r$, we have $|p(z)| \leq |p(w)|$.
If $p$ is a nonzero polynomial, then there exists a real number $r$ such that for all complex numbers $z$ with $|z| \geq r$, we have $|p(z)| \geq d$.
There exists a complex number $z$ such that for all complex numbers $w$, we have $|p(z)| \leq |p(w)|$.
If a polynomial is not constant, then its length is at least 2.
If $p$ is a polynomial, then $x^n p(x)$ is the same as $x^n$ times $p(x)$.
If $p$ is a non-zero polynomial, then there exists a non-zero constant $a$ and a polynomial $q$ such that $p(z) = z^k a q(z)$ for some $k \geq 0$.
If a polynomial is not constant, then it can be written as $p(z) = p(0) + z^k p_1(z)$, where $p_1$ is a polynomial of degree $n-k-1$ and $k$ is a positive integer.
If a polynomial $p$ is not constant, then it has a root.
If a polynomial $p$ has no constant term, then it has a root.
If $p$ and $q$ are polynomials such that $p(x) = 0$ implies $q(x) = 0$, then $p$ divides $q^n$ if $p$ has degree $n$.
If $p$ and $q$ are univariate polynomials, then $p$ divides $q^{\deg(p)}$ or $p = q = 0$.
A polynomial is constant if and only if its degree is zero.
The polynomial $0$ is the zero polynomial, the polynomial $c$ is the constant polynomial $c$, and the polynomial $x$ is the linear polynomial $x$.
The polynomial $0$ is the same as the polynomial $[0]$.
For any polynomial $p$ and $q$ and any ring element $x$, $p(x) - q(x) = p(x) + (-1)q(x)$.
If $p(x) = 0$, then $p(x) = 0$.
If $x = 0$ and $a \neq 0$, then $y = 0$ if and only if $a y - b x = 0$.
If $p$ divides $q$, then $p$ divides $p \cdot 0 + q$.
If $p$ is a nonzero polynomial and $q$ is a polynomial with degree less than the degree of $p$, then $p$ divides $q$ if and only if $q = 0$.
If $p$ divides $p'$, $q$ and $r$ are polynomials such that $aq - p' = r$, then $p$ divides $q$ if and only if $p$ divides $r$.
The following are equivalent: $\exists x. p(x) = 0 \land 0(x) \neq 0$ $\exists x. 0(x) \neq 0$ $\exists x. c(x) \neq 0$ $\exists x. 0(x) = 0$ $\exists x. c(x) = 0$
If $p$ is a non-zero polynomial, then there exists a complex number $x$ such that $p(x) = 0$.
For a polynomial $p$, the statement $\exists x. p(x) \neq 0$ is equivalent to the statement $p \neq 0$.
If $p$ is a nonzero polynomial, then there exists a complex number $x$ such that $p(x) = 0$ and $q(x) \neq 0$ if and only if $p$ does not divide $q^n$, where $n$ is the degree of $p$.
If $q^n$ and $r$ have the same values at all points, then $p$ divides $q^n$ if and only if $p$ divides $r$.
For any commutative ring $R$ and any $x \in R$, the polynomial $p(x) = c$ is equal to $y$ if and only if $c = y$.
If $f$ is a holomorphic function on a contractible set $S$ and $f(a) \neq \pm 1$, then there exists a holomorphic function $g$ on $S$ such that $f(z) = \cos(\pi g(z))$ for all $z \in S$.
For any positive integer $n$, we have $n + \sqrt{n^2 - 1} > 0$.
For any real number $x \geq 0$, there exists a positive integer $n$ such that $\left|x - \frac{\ln(n + \sqrt{n^2 - 1})}{\pi}\right| < \frac{1}{2}$.
If $z$ is a complex number of the form $m + \frac{1}{\pi} \ln(n + \sqrt{n^2 - 1})$ or $m - \frac{1}{\pi} \ln(n + \sqrt{n^2 - 1})$ for some integers $m$ and $n$, then $\cos(\pi \cos(\pi z))$ is either $1$ or $-1$.
If $f$ is a holomorphic function on the unit disk, $f(0)$ is bounded by $r$, $f$ does not take the values $0$ or $1$ on the unit disk, and $z$ is a point in the disk with norm less than $t$, then $f(z)$ is bounded by an exponential function of $t$.
For every holomorphic function $f$ defined on a neighborhood of the origin, there exists a radius $R$ such that if $f$ is nonzero and nonconstant on the disk of radius $R$, then the derivative of $f$ at the origin is less than $1$ in absolute value.
If $f$ is a holomorphic function on the complex plane such that $f(z) \neq 0$ and $f(z) \neq 1$ for all $z$, then $f$ is constant.
If $f$ is a holomorphic function on the complex plane and $f$ does not take on the values $a$ and $b$, then $f$ is constant.
If $f$ is a holomorphic function on the complex plane that is periodic with period $p \neq 0$, then $f$ has a fixed point.
If $f$ is a holomorphic function on the complex plane that is not a translation, then there exists a point $x$ such that $f(f(x)) = x$.
Suppose that for each $i$ and each sequence $r$, there exists a strictly increasing subsequence $k$ of $r$ such that $P(i, k)$ holds. If $P(i, k)$ holds for some strictly increasing subsequence $k$ of $r$, then $P(i, k')$ holds for any strictly increasing subsequence $k'$ of $r$ that is eventually equal to $k$. Then th...
If $f$ is a sequence of functions from a countable set $S$ to a normed vector space, and if the norm of $f_n(x)$ is bounded by $M$ for all $n$ and $x \in S$, then there exists a subsequence of $f$ that converges for all $x \in S$.
Suppose $\mathcal{F}$ is a sequence of continuous functions defined on a compact set $S$. If the functions in $\mathcal{F}$ are uniformly bounded and equicontinuous, then there exists a subsequence $\mathcal{F}'$ of $\mathcal{F}$ that converges uniformly to a continuous function $g$ on $S$.
Suppose $S$ is an open set in the complex plane, and $\mathcal{F}$ is a sequence of holomorphic functions on $S$ such that the range of $\mathcal{F}$ is a bounded set. Then there exists a subsequence $\mathcal{F}'$ of $\mathcal{F}$ and a holomorphic function $g$ on $S$ such that $\mathcal{F}'$ converges uniformly to $g...
If $g$ is a uniform limit of a sequence of holomorphic functions $\{f_n\}$ on an open connected set $S$, and $g$ is not constant on $S$, then $g$ is not zero on $S$.
If a sequence of holomorphic functions $\{f_n\}$ converges uniformly to a holomorphic function $g$ on a connected open set $S$, and if each $f_n$ is injective on $S$, then $g$ is injective on $S$.
Suppose $S$ is an open connected set, $w \in S$, $r > 0$, and $Y \subseteq X$ is a set of holomorphic functions on $S$ such that $h(w) \leq r$ for all $h \in Y$. Then there exists a positive number $B$ and an open set $Z$ containing $w$ such that $h(z) \leq B$ for all $h \in Y$ and $z \in Z$.