# Can a logical disjunction only connect propositions?

John, a human being, can be either dead or alive:

dead(John) ∨ alive(John)


We can then define a variable (I'm not sure if I need "element of" or "subset of" here):

x ∈ {dead, alive}
x(John)


and state that

(x = dead) ∨ (x = alive)


Can we also state that

x = (dead ∨ alive)


or is that not possible?

I'm specifically interested in this last disjunction: Can a logical disjunction only connect propositions, or can it also connect the values of a variable?

The original question, which contained a misconception, was:

In everyday speech I can say:

John is dead or John is alive.


als well as

John is dead or alive.


But it seems to me that in logics, while I can compare propositions:

(John = dead) ∨ (John = alive)


I cannot compare predicates:

John = (dead ∨ alive)


Or is the latter possible, maybe in a different notation?

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There is a great reason why mathematical language is so limited. No ambiguities. – Asaf Karagila Jan 20 '14 at 9:11

(1) In the language of first-order logic, the formation rules are entirely clear: disjunction is still a propositional connective and can only disjoin proposition-like clauses, i.e. complete well-formed formulas.

From $Dj$ and $Ak$ we can form $(Dj \lor Ak)$. And from $Dx$ and $Ax$ we can form $(Dx \lor Ax)$ and go on to form $\forall x(Dx \lor Ax)$. And so on.

We cannot form the likes of $(D \lor A)j$ or $D(j \lor k)$.

It is as simple as that. This is one important respect in which the grammar of FOL diverges from that of ordinary language, where 'or' can disjoin propositions, names, adjectives, adverbs, noun clauses, you name it ... ) You can invent formal languages which have something like this feature, but standard FOL isn't one of them. Any elementary textbook should explain this. See e.g. my Introduction to Formal Logic, §22.2.

(2) Given $Dj \lor Aj$ we cannot go on to form $x \in \{D, A\}$. For in the first $D$ has to be a predicate (if $Dj$ is to be well-formed), and in the second $D$ would have to be a term (if $\{D, A\}$ is to be well-formed).

Of course, we could introduce a term for the set of dead people, $d = \{x \mid Dx\}$ and a term for the set of living people $a = \{x \mid Ax\}$. Then given $Dj \lor Aj$ it will follow that $j \in d \lor j \in a$. But note you need $\in$ here, not identity. John is in one set or the other, not identical to one set or the other.

And you can't put $j \in d \lor a$ (to repeat, disjunction occurs between wffs, not terms). What you need, rather, is $j \in d \cup a$, where $\cup$ is set union.

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Thank you. It was the "is" in "John is dead" that made me use the "=" instead of the "∈". And your answer made me understand that I must not use intuition and everyday speech to attempt to understand logic, but need to learn at least the basic principles. My only "introduction" to formal logic was in high school, which was thirty years ago :-) – what Jan 20 '14 at 9:55
@what - there is a long discussion in mathematical logic (starting from Frege) about the multiple meanings of "is" : (i) identity, as in a definition : "a man is a human male"; (ii) predication : "Socrates is a man", i.e. $Socrates \in \{x : x$ is a man $\}$; (iii) inclusion : "man are mortal", i.e. $\{x : x$ is a man $\} \subseteq \{x : x$ is mortal $\}$. – Mauro ALLEGRANZA Jan 20 '14 at 10:41

You are not really saying

John = dead


when you say John is dead, as you are not saying "John equals death". You are saying "John has the property of dead", which can better be noted by

dead(John)


In this way, you avoid the confusion of the following reasoning:

John = dead
----------------
John = Napoleon


which is of course false. In this way, you can of course define a predicate dead_or_alive, which has the definition

dead_or_alive(x) <=> dead(x) ∨ alive(x)


and you can then say

dead_or_alive(John)

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Good answer. But according to first-order logic rules of formation, the connective $\lor$ does not "link" the two predicates $D$ and $A$; the syntax of the language says that you are connecting two formulae : $D(x)$ and $A(x)$. Of course, when you interpret them as in the example, the result is a new formula (i.e., $D$_or_$A$) "acting" as the set "union" of the two predicates. So, strictly speaking, the conncetives are operators acting on proposition (in propositional logic) or on formulae (in first-order logic): they act on expressions that are grammatically sentences (not terms). – Mauro ALLEGRANZA Jan 20 '14 at 8:15
Correction, $D$_or_$A$ is a predicate, not a formula. – 5xum Jan 20 '14 at 8:23
Thank you, @5xum. I used your answer to correct my question. – what Jan 20 '14 at 8:27
@5xum - you are right; if you enlarge the language with the new predicate $D$_or_$A$, you can obtain the desired effect. But the new predicate symbol is a predicate symbol: it does not "contain" the connective $\lor$. – Mauro ALLEGRANZA Jan 20 '14 at 9:38

In standard propositional and predicate logic, conjunction and disjunction operate only on propositions.

But there are some logicians and philosophers who sympathize with the idea that we should also be able to apply these logical operations directly to predicates.

See, for example, The Logic of Natural Language by Fred Sommers: http://books.google.co.uk/books/about/The_Logic_of_Natural_Language.html?id=1EmWQAAACAAJ&redir_esc=y

Also, Description Logics allow logical operations on predicates: http://suanpalm3.kmutnb.ac.th/teacher/FileDL/supot121255319463.pdf

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