Exists 
更多信息
- Exists[x,expr] 可以输入为
. 字符
可以用
ex
或 \[Exists] 输入. 变量
作为下标给出. - Exists[x,cond,expr] 可以输入为
. - 在 StandardForm 中,Exists[x,expr] 输出为
. - Exists[x,cond,expr] 输出为
. - Exists 可以用于诸如 Reduce、Resolve 和 FullSimplify 的函数中.
- 条件 cond 通常用于指定一个变量的域,例如 x∈Integers.
- Exists[x,cond,expr] 等价于 Exists[x,cond&&expr].
- Exists[{x1,x2,…},…] 等价于
. - Exists[x,expr] 中的
值可以视为局部的,如同在 Block 中.
范例
打开所有单元 关闭所有单元基本范例 (1)
范围 (6)
Exists[x, x ^ 2 - x + 1 == 0]用 Resolve 证明该语句为真:
Resolve[%]Exists[x, Element[x, Reals], x ^ 2 - x + 1 == 0]用 Resolve 证明该语句为假:
Resolve[%]Exists[{x, y}, x ^ 2 + y ^ 2 < 0]未指定域时,Resolve 认为不等式中的代数变量为实数:
Resolve[%]当域为 Complexes 时,允许使不等式为 True 的复数值:
Resolve[%%, Complexes]Exists[{p, q}, !Implies[p && Implies[p, q], q]]Resolve[%]如果该表达式没有直接包含变量,Exists 会自动化简:
Exists[x, y == 0]Exists[{x, y, z}, y == z]TraditionalForm 格式输出:
Exists[x, p[x]]//TraditionalForm应用 (4)
Exists[x, a x ^ 2 + b x + c < 0]Resolve[%, Reals]R1 = (2x) ^ 2 + y ^ 2 + 2 x y ≤ 1;
R2 = x ^ 2 + y ^ 2 ≤ 2;Exists[{x, y}, Element[x | y, Reals], R1 && !R2]该语句为假,因此 R1 定义的区域被包括在 R2 定义的区域内:
Resolve[%]{r1, r2} = {RegionPlot[R1, {x, -2, 2}, {y, -2, 2}, PlotPoints -> 35, PlotStyle -> Pink], RegionPlot[R2, {x, -2, 2}, {y, -2, 2}]};Show[{r2, r1}]triangle = a > 0 && b > 0 && c > 0 && a + b > c && a + c > b && b + c > a;
s = 1 / 2(a + b + c);
F = Sqrt[s(s - a)(s - b)(s - c)];
conjecture = 27(b ^ 2 + c ^ 2 - a ^ 2) ^ 2(a ^ 2 + c ^ 2 - b ^ 2) ^ 2(a ^ 2 + b ^ 2 - c ^ 2) ^ 2 ≤ (4F) ^ 6;Exists[{a, b, c}, Element[a | b | c, Reals], triangle && !conjecture]Resolve[%]acute = a ^ 2 + b ^ 2 > c ^ 2 && a ^ 2 + c ^ 2 > b ^ 2 && b ^ 2 + c ^ 2 > a ^ 2;Exists[{a, b, c}, Element[a | b | c, Reals], triangle && acute && !conjecture]Resolve[%]statement = (a && ((a && !(b || (a && c))) || b || (a && c))) || !a || !((a && !(b || (a && c))) || b || (a && c));Resolve[Exists[{a, b, c}, !statement]]它同样可以用 TautologyQ 证明:
TautologyQ[statement]属性和关系 (5)
!Exists[x, x == a]!Exists[x, Element[x, Reals], x == a]form = Exists[{x, y}, x ≠ y && a x ^ 2 + b x + c == 0 && a y ^ 2 + b y + c == 0]Resolve[form, Reals]Reduce[form, {a, b, c}, Reals]ineqs = x ^ 2 + y ^ 2 < 2 && (x - 1) ^ 2 + (y - 1) ^ 2 < 2Resolve[Exists[{x, y}, ineqs], Reals]用 FindInstance 求出一个特解:
FindInstance[ineqs, {x, y}, Reals]Exists[x, x ^ 2 + y ^ 2 + z ^ 2 == 1 && x y == z ^ 3]用 Resolve 求出语句为真的
和
的条件:
Resolve[%]用 Eliminate 解决同样的问题:
Eliminate[x ^ 2 + y ^ 2 + z ^ 2 == 1 && x y == z ^ 3, x]Reduce[Exists[y, x y == 1], x]Reduce[Exists[y, x ^ 2 + y ^ 2 ≤ 1], x, Reals]历史
2003年引入 (5.0)
文本
Wolfram Research (2003),Exists,Wolfram 语言函数,https://reference.wolfram.com/language/ref/Exists.html.
CMS
Wolfram 语言. 2003. "Exists." Wolfram 语言与系统参考资料中心. Wolfram Research. https://reference.wolfram.com/language/ref/Exists.html.
APA
Wolfram 语言. (2003). Exists. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/Exists.html 年
BibTeX
@misc{reference.wolfram_2026_exists, author="Wolfram Research", title="{Exists}", year="2003", howpublished="\url{https://reference.wolfram.com/language/ref/Exists.html}", note=[Accessed: 05-September-2026]}
BibLaTeX
@online{reference.wolfram_2026_exists, organization={Wolfram Research}, title={Exists}, year={2003}, url={https://reference.wolfram.com/language/ref/Exists.html}, note=[Accessed: 05-September-2026]}