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I have long results of a Reduce statement in the form of {(expr1) || (expr2) || (expr3) || ...}

I am looking for a way to split them into a list {expr1, expr2, expr3, ...}

Even better, I would be happy to split the internal expressions to lists at the '&&` operators as well.

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  • $\begingroup$ So how do you want a mixture of Ands and Ors to be treated? $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Jul 16, 2016 at 15:31
  • $\begingroup$ Luckily the || are outside the expressions, and internally the expressions only have &&. I would probably prefer to have it converted into some canonical form first. $\endgroup$
    – Ron
    Jul 16, 2016 at 15:35
  • $\begingroup$ Now that you bring it up: look at BooleanConvert[]. $\endgroup$
    – J. M.'s persistent exhaustion ♦
    Jul 16, 2016 at 15:41
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    $\begingroup$ In certain cases results of Reduce can be transformed into rules by using ToRules, then this post might be helpful Assign the results from a Solve to variable(s). Nevertheless you should remember that Reduce and Solve are not equivalent in general What is the difference between Reduce and Solve?. $\endgroup$
    – Artes
    Jul 16, 2016 at 15:42
  • $\begingroup$ That was helpful - thanks! (though ToRules didn't apply in this case). $\endgroup$
    – Ron
    Jul 16, 2016 at 16:17

1 Answer 1

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When the expression is inside a List use Apply at level 1

list1 = {(expr1) || (expr2) || (expr3)};

List @@@ list1

(*  {{expr1, expr2, expr3}}  *)

list2 = {(expr1) && (expr2) && (expr3)};

List @@@ list2

(*  {{expr1, expr2, expr3}}  *)

If there is no external List then just use Apply

list3 = (expr1) || (expr2) || (expr3);

List @@ list3

(*  {expr1, expr2, expr3}  *)

list4 = (expr1) && (expr2) && (expr3);

List @@ list4

(*  {expr1, expr2, expr3}  *)
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  • $\begingroup$ I found out Level does the same thing. Converting to normal form using BooleanConvert prior to applying it helped. Many thanks! $\endgroup$
    – Ron
    Jul 16, 2016 at 16:18

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