How do I push the negation symbol as far inside the expression? $(\lnot\forall n_1\in \mathbb Z)(\exists n_2\in \mathbb Z)[n_1>n_2]$ do I need to negative $(\exists n_2\in \mathbb Z)$to $(\forall \lnot n_2\in \mathbb Z)$?
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Are $n1$ and $n2$ supposed to be $n_1$ and $n_2$, two different variables? You get subscripts with an underscore, so n_1 gives $n_1$. If the subscript is mutliple characters, enclose them in braces, so x_{00} gives $x_{00}$ instead of x_00 which gives $x_00$ – Ross Millikan Sep 25 '17 at 04:41
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What can $\forall \lnot n_2 \in \mathbb Z$ possibly mean? – James Sep 25 '17 at 04:41
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I suggest you have a look at the related questions on this site - in particular this answer: https://math.stackexchange.com/a/1990294/246902 and this question: https://math.stackexchange.com/q/1095530/246902 (if you can understand the statement in the question, then you should be able to solve your problem.) – James Sep 25 '17 at 04:44
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You change the $\forall$ to $\exists$ and negate the inside. So For $P$ any statement, which may depend on $x$, we have $\lnot \forall x P \iff \exists x \lnot P$ Intuitively, if we know something is not true for all $x$, there must be some $x$ it is not true of. Similarly $\lnot \exists x P \iff \forall x \lnot P$. Intuitively, if there is no $x$ for which $P$ is true, then for all $x$ $P$ is false.
Ross Millikan
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