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I am trying match variables with subscripts just returning true or false depending if pattern is matched. My try is to use Cases. However the following does not return a match

Cases[Subscript[x, 2], Subscript[x, l_] /; l > 1]

Moreover, if this would work there must something nicer than

Length[    Cases[Subscript[x, 2], Subscript[x, l_] /; l > 1]]>=1
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  • $\begingroup$ Cases[{Subscript[x, 2]}, Subscript[x, l_] /; l > 1] , MatchQ[Subscript[x, 2], Subscript[x, l_] /; l > 1]for just single test. $\endgroup$
    – ciao
    Mar 26, 2014 at 21:46
  • $\begingroup$ MatchQ[( Subscript[x, 2])^2, _ Subscript[x, l_] _ /; l > 1] does not work, I want to check if it occurs anywhere in the term i.e. would the rule Subscript[x, l_] _ /; l > 1:> 5 change something $\endgroup$
    – warsaga
    Mar 26, 2014 at 22:05
  • $\begingroup$ That's because your use of MatchQ there is a bit wacky. Perhaps clarify more precisely what you want in the OP? It seems perhaps Cases[Level[expression, Infinity], Subscript[x, l_] /; l > 1] might be what you're after. $\endgroup$
    – ciao
    Mar 26, 2014 at 23:05
  • $\begingroup$ Not@FreeQ[(Subscript[x, 2])^2, Subscript[x, l_] /; l > 1]? $\endgroup$
    – kglr
    Mar 27, 2014 at 0:08

1 Answer 1

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As mentioned in the comments, you can use either MatchQ:

MatchQ[Subscript[x, 2], Subscript[x, _?(# > 1 &)]]

or FreeQ with Not:

!FreeQ[Subscript[x, 2], Subscript[x, _?(# > 1 &)]]
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