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Note: I'll use the convention adopted by my professor: "$\rightarrow$" is the symbol for implication and "$\implies$" for deduction.

Our professor defined implication as

The connective such that $(p\rightarrow q)$ is false only if $p$ is true and $q$ is false.

and double implication as

The connective such that $(p\leftrightarrow q)$ is true if and only if $p,q$ have the same truth value.

My problem with such definitions is that both use what they are defining: only if and if and only if are just the english expressions respectively for "$\rightarrow$" and "$\leftrightarrow$". I remember an answer here on MSE where it was indeed stated that the first one is circular, but the author somewhat accepted it. Why should we? In particular, why aren't

Implication is the connective such that $$\begin{cases} p \\ p\rightarrow q \end{cases} \implies q .$$

and

Double implication is the connective given by the conjuction of an implication and its inverse.

better?

Vincenzo Oliva
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2 Answers2

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Because you're misunderstanding the definition. (It's not your fault; it's due to the fact that the notation used by your professor is imprecise, but it is the conventional notation.)

He/she did not define the implication and equivalence symbols using themselves. Rather he defined how to interpret the truth-value of certain kinds of strings of symbols called well-formed formulae. A string like "$A \land B$" has no meaning whatsoever until it is interpreted. To interpret one has to break it down into its smallest pieces and interpret piece by piece. In propositional logic these smallest pieces are atoms, each of which is interpreted as either true or false. Then based on how the original formula is built up from atoms and logical symbols, it is recursively interpreted. The first definition you are asking about could have been stated much more clearly as:

Given well-formed formulae $p$ and $q$, the formula $p + ``\rightarrow\text{"} + q$ has truth-value $false$ if and only if $p$ has truth-value $true$ but $q$ has truth-value $false$.

Remember that formulae are just meaningless strings until interpreted! Also, "$+$" in the above denote string concatenation. Thus you can see that given such rules of interpretation of well-formed formulae, one can recursively evaluate its truth-value if one knows the truth-value of the smallest pieces.

So these definitions are not circular because we are simply defining how to interpret strings. (You can trivially write a computer program to do it too!) In short, the implication symbol is totally not the implication in the English expressions, although its interpretation as defined above is indeed intended to reflect the external concept of a conditional assertion (if $p$ is true, then $q$ is true) that is true exactly when it is not a false assertion (think of "false promise", which only occurs when $p$ is true but $q$ is false).

Now for your second question, the answer is that it is circular. I answered essentially the same question at https://math.stackexchange.com/a/1334753/21820 before. Your proposal does not at all solve the problem, because you cannot define "$\Rightarrow$" without the same circularity.

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user21820
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If you define implication as satisfying {p, (p $\rightarrow$ q)} ⟹ q, then what can get said about the truth-table for $\rightarrow$?

 → 
 p \ q  0  1
 0      
 1      0  1

But notice that we don't have to fill in the 0 row for this function. Consequently, there exist four two-valued truth functions which satisfy your rule. Consequently, if we did define implication as you proposed, it would not consist of a function in the context of two-valued logic. Also, logical equivalence wouldn't qualify as distinguishable from implication.

Thus, there is no such thing as "the connective" for what you defined.

Doug Spoonwood
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