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Exists

Exists[x,expr]

represents the statement that there exists a value of x for which expr is True.

Exists[x,cond,expr]

states that there exists an x satisfying the condition cond for which expr is True.

Exists[{x₁,x₂,…},expr]

states that there exist values for all the for which expr is True.

Details

Exists[x,expr] can be entered as . The character can be entered as ex or \[Exists]. The variable is given as a subscript.
Exists[x,cond,expr] can be entered as .
In StandardForm, Exists[x,expr] is output as .
Exists[x,cond,expr] is output as .
Exists can be used in such functions as Reduce, Resolve, and FullSimplify.
The condition cond is often used to specify the domain of a variable, as in x∈Integers.
Exists[x,cond,expr] is equivalent to Exists[x,cond&&expr].
Exists[{x₁,x₂,…},…] is equivalent to .
The value of in Exists[x,expr] is taken to be localized, as in Block.

Examples

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Basic Examples (1)

This states that there exists a positive solution to the equation :

Use Resolve to get a condition on real parameters for which the statement is true:

Reduce gives the condition in a solved form:

Scope (6)

This states that there exists for which the equation is true:

Use Resolve to prove that the statement is true:

This states that there exists a real for which the equation is true:

Use Resolve to prove that the statement is false:

This states that there exists a pair for which the inequality is true:

With domain not specified, Resolve considers algebraic variables in inequalities to be real:

With domain Complexes, complex values that make the inequality True are allowed:

This states that the negation of a tautology is satisfiable:

Use Resolve to prove it False:

If the expression does not explicitly contain the variable, Exists simplifies automatically:

TraditionalForm formatting:

Applications (4)

This states that a quadratic attains negative values:

This gives explicit conditions on real parameters:

Test whether one region is included in another:

This states that there are points satisfying R1 and not R2:

The statement is false, hence the region defined by R1 is included in the region defined by R2:

Plot the relationship:

Test geometric conjectures:

This states that there is a triangle for which the conjecture is not true:

The statement is true, hence the conjecture is not true for arbitrary triangles:

This states that there is an acute triangle for which the conjecture is not true:

The statement is false, hence the conjecture is true for all acute triangles:

Prove that a statement is a tautology:

This proves that there are no values of for which the statement is not true:

This can be proven with TautologyQ as well:

Properties & Relations (5)

Negation of Exists gives ForAll:

Quantifiers can be eliminated using Resolve or Reduce:

This eliminates the quantifier:

This eliminates the quantifier and solves the resulting equations and inequalities:

This shows that a system of inequalities has solutions:

Use FindInstance to find an explicit solution instance:

This states that there exists a complex for which the equations are satisfied:

Use Resolve to find conditions on and for which the statement is true:

This solves the same problem using Eliminate:

This finds the projection of the complex algebraic set along the axis:

This finds the projection of the real unit disc along the axis:

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