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Fix typo #34

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Fix typo #34

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2 changes: 1 addition & 1 deletion lection-10.tex
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Expand Up @@ -58,7 +58,7 @@
\begin{proof}
Пусть формальная арифметика разрешима.
Значит, есть рекурсивная функция $f(x)$: $f(x)=1$ тогда и только тогда,
когда $x \in \text{Д}_\text{Ф.А.}$. То есть, $\text{Д}_\text{Ф.А.}$ выразимо в формальной арифметике.
когда $x \in \text{Д}_\text{Ф.А.}$. То есть $\text{Д}_\text{Ф.А.}$ выразимо в формальной арифметике.

По теореме о невыразимости доказуемости,
$\text{Д}_\text{Ф.А.}$ невыразимо в формальной арифметике. Противоречие.
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