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How can I prove that given a predicate $hc$ and a one-way function $f$ that $hc$ is not hard-core? I was thinking about something like that:

i define $f(x) = (g(x), hc(x))$ that is one-way but $hc$ is not hard-core 'cause it's value is easly reconstructable from the output.

Is it right? How can I actually prove that?

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    $\begingroup$ You must exhibit an efficient algorithm that computes $\mathrm{hc}(x)$ with non-negligible probability when given $f(x)$ (for uniformly random $x$). $\endgroup$
    – fkraiem
    May 29, 2019 at 14:26
  • $\begingroup$ Do I understand you correctly, that $g$ is a OWF and $hc$ is a hardcore predicate of $g$? And your question is why $hc$ is not also a hardcore predicate of $f$ as defined in the question? $\endgroup$
    – Maeher
    May 29, 2019 at 19:43
  • $\begingroup$ $G$ is a OWF and given the function $f$ that is something like $f(x) = (g(x), hc(x))$ I have to prove that eventhogh $f$ is OWF, $hc$ is not hard core for $f$ $\endgroup$
    – quaqua
    May 30, 2019 at 14:59
  • $\begingroup$ @quaqua: So like given $(g(x), hc(x))$ make a good guess at the value of $hc(x)$? $\endgroup$
    – Ry-
    May 31, 2019 at 0:59
  • $\begingroup$ @Ry yes, i have to prove it somehow $\endgroup$
    – quaqua
    May 31, 2019 at 12:10

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