How would a statement beginning, "For all $x$ in set $S$ except $x=a$..." be translated into symbolic predicate logic? I'm somewhat of a purist in symbols and am less than satisfied with "$\forall x \in S$ except $x = a$..." or "$\forall x \in S$ where $x \ne a$...".
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1$\begingroup$ As the comment on one of the answers suggests, the question in the title is not the same as the question in the body. The title says "except one...", which may be any one, and need not be a in particular: it might be b or c or .... It is more complicated to express "all elements except exactly one" than it is to express "all except this particular one." $\endgroup$– MikeCCommented Jan 12, 2012 at 4:48
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3 Answers
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The exact correspondence in first order language is as follows:
$∀x((x∈S ∧ x≠a) → ⋯)$
It reads: for all $x$, if $x$ belongs to $S$ and $x$ is not $a$, then....
$x≠a$ is just a short form for $￢(x=a)$
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2$\begingroup$ This is good if we interpret "for all $x$ except $a$, $Q(x)$ holds" as meaning "for all $x$ except possibly $a$, $Q(x)$ holds." If we interpret "except $a$" as implying that $Q(a)$ fails, a different translation is needed. $\endgroup$ Commented Jan 11, 2012 at 8:13
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$\begingroup$ @AndréNicolas Which translation do you suggest? $\endgroup$– magmaCommented Jan 11, 2012 at 11:19
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$\begingroup$ @AndréNicolas "Except possibly a" is modal logic I believe. The OP asked for first-order logic. $\endgroup$– magmaCommented Jan 11, 2012 at 11:26
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$\begingroup$ I don't modal. Example: "Everyone in this room except Alice is overweight." Your translation says that if $x$ is not Alice, then $x$ is overweight. Logically speaking, it says nothing about Alice's weight. That may be what is intended by the sentence "Everyone in this room except Alice is overweight." But probably not. To be more mathematical, "Every real number except $0$ has a multiplicative inverse." If by this sentence we mean in particular that $0$ doesn't have a multiplicative inverse, your translation does not do the job. If we mean to say nothing about $0$, it does. $\endgroup$ Commented Jan 11, 2012 at 16:32
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$\begingroup$ @AndréNicolas you are right, my PARTIAL statement (it has dots "...") does not say anything about a, that is because the OP did not say anything about a. You seem to suggest that what goes for the others, does not go for a (that is: it is negated). But that it's not always so. Depends on the rest of the sentence (the dots). In the dots you fill what happens to a. For example: "Everyone in this room except Alice is overweight and diabetic, Alice is just overweight". In the dots you put: Ox ∧ Dx ∧ ￢Oa. $\endgroup$– magmaCommented Jan 12, 2012 at 4:39
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I would use $(\exists a\in S)(\forall x\in S)(x\neq a \Leftrightarrow \cdots)$. It's not a direct correspondence to the English, but not everything in symbolic logic is.
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4$\begingroup$ this is not the correct translation I am afraid. It does admit the existence or possibly many a. (it says there is at least one a such that...). While the English text only admits one element called a. Syntactically you are replacing a constant (a) with a bound variable (also called a in your formula). But a bound variable is a dummy variable, so you can call it anything, for example y. And then it loses its unique identity $\endgroup$– magmaCommented Jan 11, 2012 at 1:37
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$\begingroup$ @magma: I think jwodder was answering the question in the subject as opposed to the question in the text; the text mentions a specific $a$, but I also initially read the question as asking about 'for all but one (unknown) element of the set $S$' from its subject. $\endgroup$ Commented Jan 11, 2012 at 2:08
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$\begingroup$ @StevenStadnicki : jwodder statement says: for all x in S except SOME....so it does not reflect neither the title nor the body of the question. Besides, the title says exactly the same thing as the body (except one means except only one) $\endgroup$– magmaCommented Jan 11, 2012 at 2:43
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2$\begingroup$ I think this is a valid interpretation of the title of the question. $\endgroup$ Commented Jan 12, 2012 at 0:00
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$\begingroup$ More precisely, it is a valid interpretation of the title the question had when this answer was written (and still had when I wrote my comment). Then it said: "for all elements in a set except one..." $\endgroup$ Commented Jan 15, 2012 at 2:27