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# ForAll ()

 ForAll[x, expr]represents the statement that expr is True for all values of . ForAll[x, cond, expr]states that expr is True for all x satisfying the condition cond. ForAll[{x1, x2, ...}, expr]states that expr is True for all values of all the .
• ForAll[x, expr] can be entered as . The character can be entered as Esc fa Esc or \[ForAll]. The variable is given as a subscript.
• ForAll[x, cond, expr] can be entered as .
• ForAll[x, cond, expr] is output as x, condexpr.
• The condition cond is often used to specify the domain of a variable, as in .
• ForAll[{x1, x2, ...}, ...] is equivalent to .
• The value of in ForAll[x, expr] is taken to be localized, as in Block.
This states that for all , is positive:
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 for all , is positive:
 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 for all the inequation is true:
Use Resolve to prove that the statement is false:
This states that for all real the inequation is true:
Use Resolve to prove that the statement is true:
This states that for all pairs 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 False are allowed:
This states the tautology implies :
Prove it:
If the expression does not explicitly contain a variable, ForAll simplifies automatically:
 Applications   (5)
This states the inequality between the arithmetic mean and the geometric mean:
Use Resolve to prove the inequality:
This states a special case of Hölder's inequality:
Use Resolve to prove the inequality:
This states a special case of Minkowski's inequality:
Use Resolve to prove the inequality:
Prove geometric inequalities for , , and sides of a triangle:
This states that an inequality is satisfied for all triangles:
Use Resolve to prove the inequality:
This states that an inequality is satisfied for all acute triangles:
Use Resolve to prove the inequality:
Test whether one region is included in another:
This states that all points satisfying R1 also satisfy R2:
The statement is true, hence the region defined by R1 is included in the region defined by R2:
Plot the relationship:
Negation of ForAll gives Exists:
Quantifiers can be eliminated using Resolve or Reduce:
This eliminates the quantifier:
This eliminates the quantifier and solves the resulting equations and inequalities:
This states that an equation is true for all complex values of :
Use Reduce to find the values of parameters for which the statement is true:
This solves the same problem using SolveAlways:
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