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I am trying to prove that all statements are logically equivalent to a statement that does not use negation and in which the only logical connective used is nand (defined as $\neg(p\ \land\ q)$ for statements $p,q$).

In brief, the way I am doing it is with two cases:

Let $S$ be a statement. If $S$ is atomic, then we are done (the statement that does not use negation or logical connectives other than nand with which $S$ is logically equivalent is $S$ itself). If $S$ is non-atomic, then we simply have to show that all possible mappings of $(p,q)$ to the boolean domain (there are 16 of them) can be represented using only nand, where $p\ \otimes\ q \Leftrightarrow S$ and $\otimes$ is some logical connective.

The part I am unsure of is whether it is safe to jump to the idea that $S$ can always be represented as $p\ \otimes\ q$. It seems natural that if $S\ \Leftrightarrow\ (p\ \otimes\ q)\ \otimes\ r$ (the two instances of $\otimes$ need not necessarily refer to the same logical connective), for example, then you would apply the arguments in the second case twice: first to part in parentheses, and then to the whole expression.

How might I convey this idea of recursion in a rigorous proof? Do I even need to? I.e. is it implied?

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Take your favorite set of complete logical connectives. {AND, OR, NOT} obviously works, but so does {AND, NOT} and {OR, NOT} if you want to do a simple lemma invoking De Morgan's laws. Then show that every member of that set can be expressed with just NAND.

Then you can let $S$ represent the set of boolean expressions that are equivalent to an expression just using literals and NAND as a connector. You know that all literals are in $S$ by definition. And you know that if $A\in S$ and $B\in S$, then $A\land B\in S$, $A\lor B\in S$, and $\neg A\in S$, by what you proved in the previous paragraph and the induction hypothesis. Therefore, by induction on the construction of boolean expressions, $S$ represents all boolean expressions.

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  • $\begingroup$ Thanks, but I am trying to avoid relying on knowing the functional completeness of a different set of logical connectives. $\endgroup$
    – RobertR
    Commented Sep 14, 2019 at 21:33
  • $\begingroup$ @RobertR Okay, then you'll have to do it for all of the connectors. It isn't too rough, NOT is the simplest and then AND and then you can substitute into the other ones based on the identities/definitions you know. $\endgroup$
    – user694818
    Commented Sep 14, 2019 at 21:36

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