... false.^2.1

Recall that a prime number is one that is divisible only by itself and 1.

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... theory.^2.2

Goldbach famously conjectured that the answer is ``Yes."

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... objects.^2.3

This is the so-called ``naive" conception of a set -- naive because, as Bertrand Russell proved to Gotleib Frege's horror, this conception quickly leads to outright contradiction. (Russell's proof has come to be known as Russell's Paradox.) There are ways to circumvent this contradiction (via, e.g., axiomatic set theory, one example of which is the Zermelo-Fraenkel approach), but these routes are outside our purposes in the present chapter. Russell's Paradox and axiomatic set theory are discussed in the chapter ``Set Theory" in this book.

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... disjoint.^2.4

That is, two sets A and B are disjoint if and only if $A \cap B = \emptyset$.

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... alone.^3.1

To make it absolutely clear that one can infer to either conjunct we could include the following slight variant as covered under this rule.

$$\begin{array}{r\vert ll} \vdots & \vdots & \vdots\\ k & \ph... ... \wedge \mbox{ Elim}\\ \vdots & \vdots & \vdots\\ \end{array}$$

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... schema:^3.2

Actually, the negative and positive can be reversed in this scheme. That is, the lines after the first clause can be positive, and can clash with negative disjunts in the first line.

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...$_{PC}, \phi \in \Gamma^\star \mbox{ iff } \neg\phi \in \Gamma^\star$.^3.3

To prove (++) true for the remaining four forms, the corresponding facts are needed. For example, the proposition corresponding to conjunction is

($\ast^\wedge$)

For all formulas $\phi, \psi \in \cal L$ $_{PC}, (\phi \wedge \psi) \in \Gamma^\star \mbox{ iff } \phi \in \Gamma^\star \mbox{ and } \psi \in \Gamma^\star.$

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...-sentence^4.1

A sentence is a formula that contains no free variables.

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... substitution:^4.2

By tradition, the ordered pairs are represented with the variable appearing to the left of the slash /, and the term appearing to the right of this symbol.

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...$\phi$.^4.3

Can you figure out how to define the concept of rank? If $\phi$ is atomic, then rank($\phi$) is 0; and then for every connective we increase the rank by 1. So, e.g., we say that the rank of $\neg \psi$ is 1 plus rank$(\psi)$. Can you formalize the entire inductive definition?

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