The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element.  Alternatively you may use op(i,S).  The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used.  S[-1] selects the last element, S[-2] the second last, and so on.

Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op(i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used.  Thus S[2..-2] selects all elements except the first and last.

Lists and sets can be nested, in which case selection can be done in one of two ways:  S[i][j]...[n] or S[i,j,...,n].

To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[].

Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup \left\{x\right\}$).

Replacing the i-th element of a list L by x can be done by subsop(i=x, L).

The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small.

Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S\setminus \left\{x\right\}$).

To test if an element x is in a list or set L, use either member(x,L) or x in L ($x\in L$).

To test if a set S is a subset of a set T, use the subset operator  S subset T  ($S\subseteq T$).

To find the intersection of two sets S and T, use S intersect T ($S\cap T$).

To find the union of two sets S and T, use S union T ($S\cup T$).

To find the difference of two sets S and T, use S minus T ($S\setminus T$).

For more information on these set operations, see Set Operators.

Further list operations can be found in the package ListTools.

object id: this causes the same kind of data-structures to be grouped together.  For example integers will come before exact rationals, floats, and complex numbers.

object length: this causes objects of the same size to be grouped together.  The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first.

lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order.

recurse on components: this causes compound objects to be ordered according to the above properties of their child components.  For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length.  So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3.  Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y.

address: in some cases the object will be identical in every way, such as the local name x and the global name x.  When this happens the machine address of the given object will be used.  This at least ensures the set order will be consistent within the running session.

{x,y,y};

$\left\{x\,y\right\}$

{y,x,y};

$\left\{x\,y\right\}$

{B,b,c,a,A};

$\left\{A\,B\,a\,b\,c\right\}$

S:={5,4,3,2,1};

$S≔\left\{1\,2\,3\,4\,5\right\}$

$S\mathbf{union}\left\{3\,6\right\}$

$\left\{1\,2\,3\,4\,5\,6\right\}$

$S$

$\left\{1\,2\,3\,4\,5\right\}$

Set operators do not work on lists.

$\left\{1\,2\,3\right\}\mathbf{intersect}\left\{2\,3\,4\right\}$

$\left\{2\,3\right\}$

$\left[1\,2\,3\right]\mathbf{intersect}\left\{2\,3\,4\right\}$

Error, invalid input: intersect received [1, 2, 3], which is not valid for its 1st argument

$\mathrm{evalb}\left(1\mathbf{in}\left\{1\,2\,3\right\}\right)$

$\mathrm{true}$

$\mathrm{evalb}\left(4\mathbf{in}\left[1\,2\,3\right]\right)$

$\mathrm{false}$

$S≔\left\{1\,2\right\}$

$S≔\left\{1\,2\right\}$

$T≔\left\{1\,2\,3\right\}$

$T≔\left\{1\,2\,3\right\}$

$S\mathbf{subset}T$

$\mathrm{true}$

$T\mathbf{minus}S$

$\left\{3\right\}$

$\left[x\,y\,y\right]$

$\left[x\,y\,y\right]$

$\left[y\,x\,y\right]$

$\left[y\,x\,y\right]$

$L≔\left[\mathrm{seq}\left(x\left[i\right]\,i=1..4\right)\right]$

$L≔\left[x_1\,x_2\,x_3\,x_4\right]$

$\mathrm{numelems}\left(L\right)$

$4$

$L\left[2\right]$

$x_2$

$L\left[1..2\right]$

$\left[x_1\,x_2\right]$

$L\left[2..-1\right]$

$\left[x_2\,x_3\,x_4\right]$

$\mathrm{op}\left(L\right)$

$x_1,x_2,x_3,x_4$

$L\left[\right]$

$x_1,x_2,x_3,x_4$

$\mathrm{op}\left(L\left[2..3\right]\right)$

$x_2,x_3$

$\mathrm{op}\left(2..3\,L\right)$

$x_2,x_3$

$L≔\left[\mathrm{op}\left(L\right)\,x\left[5\right]\right]$

$L≔\left[x_1\,x_2\,x_3\,x_4\,x_5\right]$

$\mathrm{subsop}\left(2=\mathrm{NULL}\,L\right)$

$\left[x_1\,x_3\,x_4\,x_5\right]$

$\mathrm{member}\left(x\left[1\right]\,L\right)$

$\mathrm{true}$

$L≔\left[1\,\left[2\,3\right]\,\left[4\,\left[5\,6\right]\,7\right]\,8\,9\right]$

$L≔\left[1\,\left[2\,3\right]\,\left[4\,\left[5\,6\right]\,7\right]\,8\,9\right]$

$L\left[3,2,1\right]$

$5$

$L\left[3\right]\left[2\right]\left[1\right]$

$5$