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# Exists ()

 Exists[x, expr]represents the statement that there exists a value of for which expr is True. Exists[x, cond, expr]states that there exists an 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.
• Exists[x, expr] can be entered as . The character can be entered as Esc ex Esc or \[Exists]. The variable is given as a subscript.
• Exists[x, cond, expr] can be entered as .
• Exists[x, cond, expr] is output as .
• The condition cond is often used to specify the domain of a variable, as in .
• 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.
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:
This states that there exists a positive solution to the equation :
 Out[1]=
Use Resolve to get a condition on real parameters for which the statement is true:
 Out[2]=
Reduce gives the condition in a solved form:
 Out[3]=
 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:
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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