When applying Beta reduction does the function also affect on the $\lambda$ term? (If same value)

For example

$\lambda$ z.$\lambda$ z (z z) t

What is the correct reduction?

$\lambda$z (t t) or

$\lambda$t (t t)

  • 1
    $\begingroup$ How is λz.λz (z z) t parsed? Is it λz.(λz (z z) t) or (λz.λz (z z)) t? $\endgroup$ – Daniel Fischer Jun 23 '15 at 13:55
  • $\begingroup$ it is (λz.λz (z z)) t $\endgroup$ – janitha000 Jun 26 '15 at 13:03
  • $\begingroup$ Then you get $\lambda u\, (u\; u)$. Of course instead of $u$, you can call the variable anything you like, e.g. $t$ or $z$. Note that $\lambda z.\, \lambda z\, (z\; z) \equiv \lambda z.\, \lambda y\, (y\; y)$. $\endgroup$ – Daniel Fischer Jun 26 '15 at 13:07

The beta contraction of $(\lambda x . M)\, N$ is defined as $[N/x]\, M$, and recall that the substitution is done by replacing every free instance of $x$ by $N$.

Explicitly, $(\lambda z. (\lambda z. zz))t \triangleright_\beta [t/z]\,(\lambda z. zz)$, but as there is no free occurence of $z$ no substitution is performed.


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