It is best explained using "deparse with extra parentheses":

% perl -MO=Deparse,-p -ce '($x) ? $z  = "true" : $z  = "false";'
(($x ? ($z = "true") : $z) = "false");
-e syntax OK
%
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That's a lot of parens, the important pair is: ($x ? $z = "true" : $z) = "false".

The precedence is such that in this case, the ternary expression is $x ? $z  = "true" : $z. Let's call the result of that $y. That then forms the left hand side of $y = "false". So when $x is true it assigns "true" to $z, then yields $z as an lvalue to apply the second assignment $z = "false", immediately overwriting the "true".

The moral of the story is: put disambiguating parens round all but the simplest expressions in a ternary operator.

% perl -wle '$x = 1; $x ? ($z = "true") : ($z = "false"); print $z'
true
% # or of course make them simple expressions:
% perl -wle '$x = 1; $z = $x ? "true" : "false"; print $z'
true
%
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