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    $\begingroup$ If you look at FullForm you'll see that you need {Sqrt[(t_)^2] -> t, 1/Sqrt[t_^2] -> 1/t}. $\endgroup$ Commented Nov 5, 2013 at 17:09
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    $\begingroup$ Yep, always use FullForm[] when a replacement doesn't work $\endgroup$ Commented Nov 5, 2013 at 17:11
  • $\begingroup$ It worked. But I don't understand the concept. Let's say, we want to replace E^(1 - x[1] + x[2]) to m[1,2]. Then the replacement doesn't change E^(2 - 2 (x[1] - x[2])) even thought it should be replaced by m[1,2]^2. $\endgroup$ Commented Nov 5, 2013 at 17:27
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    $\begingroup$ Because FullForm[Sqrt[x]] == Power[x, Rational[1,2]] while FullForm[1/Sqrt[x]] == Power[x, Rational[-1,2]] and therefore your pattern will not match. $\endgroup$ Commented Nov 5, 2013 at 17:58
  • $\begingroup$ This question has an answer here: Pitfalls. Related question: Why is ReplaceAll behaving like this? $\endgroup$ Commented Nov 5, 2013 at 19:02