The apply(p, rest) calling sequence returns the expression $p\left(\mathrm{rest}\right)$.

Using unapply, the apply procedure satisfies the functional equation apply(unapply(p, v), op(v)) = p.

If p has special evaluation rules, then these rules are not respected.

apply(g);

$g\left(\right)$

apply(f, s);

$f\left(s\right)$

apply(f, s, t, u, v);

$f\left(s\,t\,u\,v\right)$

apply(sin, Pi);

$0$

map(apply, [sin, cos, tan], Pi);

$\left[0\,\mathrm{-1}\,0\right]$

apply( [sin, cos, tan], Pi );

$\left[0\,\mathrm{-1}\,0\right]$

apply(unapply([x^2,exp(x),x],x),2);

$\left[4\,{\ⅇ}^2\,2\right]$

apply(unapply(x^2 + y^2, [x, y]), a, b);

$a^2+b^2$

map(proc(n)
    description "an anonymous and recursive procedure";
        (x -> apply(x, x))(f -> proc(p, c)
                                    if c > n then
                                            p
                                    else
                                            (f(f))(p * c, 1 + c)
                                    end if
                                end proc)(1, 1)
    end proc, [$ 1 .. 10]);

$\left[1\,2\,6\,24\,120\,720\,5040\,40320\,362880\,3628800\right]$

Typesetting[Typeset](sqrt(x));

$\sqrt{x}$

lprint(Typesetting[Typeset](sqrt(x)));

Typesetting:-msqrt(Typesetting:-mi("x"))

x := 5;

$x≔5$

Typesetting[Typeset](sqrt(x));

$\sqrt{x}$

apply(Typesetting[Typeset], sqrt(x));

$\sqrt{5}$