Looking at perlop, you will see that lc, a named unary operator, is lower in precedence than the binding (pattern match) operator, "=~". So your second example is actually evaluated as: 
if (lc ($foo =~ (lc $bar))

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Operator precedence can be a tricky thing, and it usually bites you in the posterior just when you think you've got it down. 
   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

In reply to Re: Operator precedence by MeowChow
in thread Operator precedence by Kickstart

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