[expr] bit-wise "or" operator

Arguments must be integers, result is an integer.

Bit ''n'' of the result is 0 if bit ''n'' of each argument is 0. Otherwise, bit ''n'' of the result is 1.

For negative arguments, observe that ~$n==(-1-n) (See the [~] operator for further discussion.)  The following cases then exist:

  Case            Result
  ---------------------------------------------------------------------------
  $a>=0, $b>=0    $a|$b, as defined above.

  $a>=0, $b<0     $a|$b == ~(~$a & ~$b)        De Morgan's Law
                        == ~(~$a & (-1-$b))    Extended definition of [~]
                        == -1-(~$a & (-1-$b))  Extended definition of [~]
                  The expression (-1-$b) is nonnegative, and so the 
                  expression (~$a & (-1-$b)) can be evaluated by bitwise
                  operations.

  $a<0, $b>=0     Commute to ($b | $a) and solve as above.

  $a<0, $b<0      $a|$b == ~(~$a & ~$b)           De Morgan's Law
                        == -1-((-1-$a) & (-1-$b)) Extended definition of [~]
                  Since (-1-$a) and (-1-$b) are both nonnegative, the &
                  in the expression above can be evaluated wuth bitwise
                  operations. 

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