Conditional or biconditional for 'except'?

Question

The following translation is given in (part (c) of Q1 of Exercise 2.1 in) the book How to Prove It by Daniel Velleman:

Everyone likes Mary, except Mary herself.

$$\forall x(\neg(x=m)\rightarrow L(x, m)),$$

where $L(x, y)$ stands for "$x$ likes $y$" and $m$ for "Mary".

Acording to this formalisation, $L(m, m)$ is possible. But, according to given question, Mary doesn't like herself, so I think that a biconditional should be used instead of a conditional. If am wrong, please explain it to me.

Answer 1

Everyone likes Mary, except Mary herself.[...] I think that a biconditional should be used instead of a conditional.

I agree.

A straightforward interpretation of this sentence is that Mary does not like herself, i.e. we should have $\neg L(m,m)$.

The conditional, by itself, leaves $L(m,m)$ undefined. I suggest the translation:

$~~~~~~~ \forall x (\neg L(x,m) \iff x=m)$

Where the domain of discussion is some group of people.

For $x=m$, we have $\neg L(m,m)$ as required.

For $x\neq m$, we have $L(x,m)$ as required.

Answer 2

There is nothing wrong in the given formula; the trouble is at its semantic evaluation. The crucial detail missing is that formulas involving identity cannot be treated in the ordinary way as those involving non-logical predicates: Identity is a dyadic logical predicate governed by its own inference rules.

The exclusive logical status of identity is a reflection of the peculiarity of the concept of identity (one may wish to take a look at the SEoP article "Identity" by Noonan and Curtis to have an overview of the relevant conceptual aspects). It should be remarked that in some texts on logic, the term 'equality' is used instead of 'identity' (see, for example, the section 2.8 First-Order Theories with Equality in Mendelson's Introduction to Mathematical Logic 6th edition), but in purely logical contexts, the term 'identity' is preferable, for we do not take into account any mathematical operations, but being numerically one (i.e., logical identity).

Notice the identity predicate is prone to interfere with the universe of discourse. Suppose we have a universe with two elements $\{a, b\}$. However, a statement $a = b$ reduces the universe virtually to one element. Also, though nothing paradoxical hangs over satisfying $x = x$, recall that $I =\{x\,\vert\,x=x\}$ is not a set by the standard conception, thus, any reference to such a "set" $I$ is fallacious. All in all, it should be clear that observance of the inference rules for identity is a sine qua non; otherwise, fallacies easily ensue.

The inference rules are formulated with some variance according to the features of logical systems. In essence, we have two rules:

$$\frac{}{t=t}\tag{1}$$

$$\begin{array}{l} t_{1}=t_{2}\tag{2}\\ \phi(t_{1})\\ \hline \phi(t_{2}) \end{array}$$

or,

$$t = t\tag{1}$$ $$t_{1} = t_{2}\rightarrow (\phi(t_{1})\rightarrow\phi(t_{2}))\tag{2}$$

where $t, t_{1}, t_{2}$ are closed terms (viz., terms with no free variables, e.g., individual constants). It is important to note that the rules (2) require an antecedent $t_{1}=t_{2}$.

Let us see how these rules unfold with the method of analytic tableaux on the question. To be more illustrative, first, we do with an arbitrary constant $\alpha$ different from $m$ in the premiss. Hence, we examine the argument

$$\forall x(\neg\alpha =m\rightarrow L(\alpha,m))\quad\therefore\,L(\alpha, m)$$

A sketch of the tree can be given as follows (with universal instantiation to $\alpha$):

$$\neg\alpha=m$$ $$\neg\alpha =m\rightarrow L(\alpha,m)$$ $$\neg L(\alpha, m)$$ $$\diagup\qquad\diagdown$$ $$\alpha =m\qquad L(\alpha,m)$$ $$\times\qquad\qquad\times$$

Both branches are closed, the argument is valid. However, we cannot proceed in the case of $m$, for we cannot take $\neg m=m$ as a premiss, which is a blunt contradiction that trivialises the whole logical enterprise. Then,

$$m=m$$ $$\neg m=m\rightarrow L(m,m)$$ $$\neg L(m, m)$$ $$\diagup\qquad\diagdown$$ $$m=m\qquad L(m,m)$$ $$\qquad\qquad\qquad\times$$

The left branch cannot be closed, the argument is invalid.

For a fairly extensive discussion of the "identity calculus", I recommend the chapter 6, "'IS' (in one sense)", of Logic: Techniques of Formal Reasoning by Kalish and Montague.