% EXERCISE TWO
\documentclass{amsart}
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\textbf{Exercise Two: Math Blackboard, Logic, Quantifiers and Alignment}
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The proposition $(P\vee Q)\vee(\mathord{\sim}P \wedge \mathord{\sim}Q)$ 
is a tautology because it is always true.\\

[\texttt{To typeset some of the following symbols, you may need to use
the Help menu. Also, try using the }\verb!\not! \texttt{ command}]

\begin{align*}
\mathord{\sim}[(\exists n\in\mathbb{N})(\forall m\in\mathbb{N})(n\leq m)] 
&\iff (\forall n\in\mathbb{N})\mathord{\sim}[(\forall m\in\mathbb{N})(n\leq m)]\\
&\iff (\forall n\in\mathbb{N})(\exists m\in\mathbb{N})\mathord{\sim}(n\leq m)\\
&\iff (\forall n\in\mathbb{N})(\exists m\in\mathbb{N})(n\not\leq m)
\end{align*}

[\texttt{Typeset the following. Then decide if the statements are equivalent.}]
$$(\forall x\in\mathbb{R})(\exists y\in\mathbb{R})(x=-y)$$
$$(\exists y\in\mathbb{R})(\forall x\in\mathbb{R})(x=-y)$$

% The statements are not equivalent. The first says correctly that for any chosen x
%   we can find a y that's the opposite of that particular x.
% The second claims falsely that there is a single y that works for all x. 

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