Often I have to do perform a succession of substitutions on some expression, and in a notebook the way I know to do that is

((((expr /. subst1) /.subst2) /.subst3) ...

So everytime I want to add a substitution I have to go back at the beginning of the line and add a ( and then go at the end of the line and add a ) and then put the new substitution. Is there a more comfortable way to do that? Something like

expr // function1 // function2 // function3 ...

On the same line, when I have to both apply functions and substitution I have to use again the brackets, because

expr // function1 /. subst1 // function2...

mess up the order in which things are done: function2 is applied to subst1 which is then applied to function1 which is then applied to expr.

  • 2
    $\begingroup$ ReplaceAll has an operator form. expr // ReplaceAll[expr -> 1] // ReplaceAll[1 -> 2] $\endgroup$
    – Kuba
    Oct 24, 2018 at 12:03
  • 3
    $\begingroup$ You don't need any of those parentheses, since /. is already left associative. Just use expr /. subst1 /.subst2 /.subst3. $\endgroup$
    – Carl Woll
    Oct 24, 2018 at 14:25
  • 2
    $\begingroup$ you might find Fold convenient: e.g., substitutions = {a -> 1, b -> d, c -> 2, d -> 3}; Fold[ReplaceAll, {a, b, c, d}, substitutions] . $\endgroup$
    – kglr
    Oct 24, 2018 at 16:50
  • 1
    $\begingroup$ Note that replacement can take list arguments, so if the substitutions are independent you could write expr/.{subst1,subst2,subst3}. It they are not independent, expr//.{subst1,subst2,subst3} might give you the answer you need. $\endgroup$
    – mikado
    Oct 24, 2018 at 17:12

1 Answer 1


Summing up, this is perfectly fine:

2 /. 2->{2,3} /. 2 -> 4

{4, 3}

And if you need to mix it with functions you can use an operator form of ReplaceAll:

4 // Sqrt // ReplaceAll[2 -> 5]



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