John Beverley

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Logic Notes 11-4 to 11-8

*I’ve pulled together notes from David Braun’s symbolic logic course at UB and Gary Hardegree’s text Symbolic Logic, which can be found online*

We can describe the meanings of the quantifiers roughly as follows (I use “V” for universal and “E” for existential in what follows):

V: for all

E: for some

Recall, the meaning of connectives like “/\” is given by their truth tables.  ‘->’ means just what the truth table says.  This may not be quite what ‘if then’ means in English, but it is close enough so that we can use “->” to symbolize ‘if then’.  We can give an analogous official explanation of the official meanings of "V” and “E”.  The official explanation gives the official meaning by saying when universal generalizations and existential generalizations are true.  

(Informal) Definitions of the quantifiers

To tell whether or not a sentence of the form Vx(...x...x...) is true:

Remove the initial universal quantifier. Pretend that the variable it was binding is a name letter. If you now have a sentence that is true no matter what the pretend constant stands for, then the original sentence is true; otherwise it is false.

To tell whether or not a sentence of the form Ex(...x...x...) is true:

Remove the initial existential quantifier. Pretend that the variable it was binding is a name letter. If there is something that the pretend constant could stand for such that the sentence you now have is true, then the original sentence is true; otherwise it is false.

Now let’s work on translations into natural language.

Symbolization and Translations

Knowing how to symbolize these four basic categorical forms will get you a long way in symbolization.

All F are G                              Vx(Fx -> Gx)

Some F are G                        Ex(Fx /\ Gx)

Some F are not G                Ex(Fx /\ ~Gx)

No F are G                              ~Ex(Fx /\ Gx) or Vx(Fx -> ~Gx)

For example:

All cats are mammals.

For all x, if x is a cat, then x is a mammal.

Vx(Cx -> Mx).

Some dogs bark.

For some x, x is a dog and x barks.

Ex(Dx /\ Bx)

Some cats are not white.

For some x, x is a cat and x is not white.

Ex(Cx /\ ~Wx)

No cats are dogs.

It is not the case that there is a cat that is a dog.

It is not the case that for some x, x is a cat and x is a dog.

~Ex(Cx /\ Dx)

Every cat is a non-dog.

Vx(Cx -> ~Dx)

Variants on “all” and “every”

All Fs are Gs.

Every F is a G

Each F is a G

All of the preceding are to be translated as:

Vx(Fx -> Gx)

The Domain of Quantification and absolutely universal quantification

Sometimes the way in which we symbolize a universal sentence will depend on the domain of quantification we choose.  Consider the following sentence.

Everyone is happy.

This will depend partly on what we take our domain of quantification to be.  It seems that ‘everyone’ here is quantifying over people.  So suppose we let the domain of quantification be just people.  Then we should symbolize as follows.

D: people

Hj: j is happy

VxHx

If we change the domain to all living things, then we need to change the symbolization.

D: all living things.

Hj: j is happy

Pj: j is a person.

Vx(Px -> Hx)

Pay attention to the domain of quantification when symbolizing. 

“Not” and ambiguity

We need to be careful about negation and the universal quantifier.  First consider the following sentences.

Not every dog is brown.

Not all dogs are brown.

These are unambiguous.  It is simply the denial of “every dog is brown”.

~Vx(Dx -> Bx)          

It is not the case that all dogs are brown.

This will be true if some dogs are brown, as long as some dogs are not brown.  The next sentence, however, is ambiguous.

All dogs are not brown.

On one way of understanding it, it simply denies that all dogs are brown.  On this way of understanding it, we symbolize it in exactly the way above.  But there is another way of understanding it: as asserting that every dog is such that it is non-brown, or as asserting that no dog is brown.  We can paraphrase these two ways of understanding the sentence as follows.

It is not the case that all dogs are brown.

All dogs are non-brown (things).

We symbolize these two readings as follows.

~Vx(Dx -> Bx)           It is not the case that all dogs are brown.

Vx(Dx -> ~Bx)           All dogs are non-brown

An unambiguous way to express the latter in English is: No dog is brown.

More on Existential Sentences

There are several variants of existential generalization in English.

Some F is G.

Some Fs are Gs.

At least one F is G.               

There is an F that is G.                                

There exists an F that is G.                             

All of the preceding are to be translated as:

Ex(Fx /\ Gx)

We may simply want to assert the existence of something.

There is an F.                          

There is a cat.

These are to be translated as:

ExFx

Existentials and negations

Some sentences containing the existential also contain a negation, as does one of our basic categorical forms.

Some F is not G

There is an x such that (x is F and x is not G).

Ex(Fx /\ ~Gx)

Watch where the negation is placed.

It is not the case that some F is G.

It is not the case that there is an x such that (x is F and x is G)

~Ex(Fx /\ Gx)

We might restate this “No F is G”, which can be symbolized in two equivalent ways.

No F is G.

~Ex(Fx & Gx)

Vx(Fx -> ~Gx) 

Existentials with arrow

Notice that the main connective used in symbolizing an existential English sentence is “/\”

and not “->”.  Here’s why.  Consider the sentence:

Some cats are dogs.

This sentence is obviously false.  Now consider the following bad symbolization of it.

Ex(Cx -> Dx)

This sentence is true!  For notice that it will be true if there is just one thing that makes conditional true.  And the conditional will be true when the antecedent false.  So if there is just one thing that is not a cat, then this symbolization will be true.  But there is: Fido, for example, is not a cat.

Another way to see this: notice that the conditional inside the formula is equivalent to a disjunction.

Ex(Cx -> Dx)

Ex(~Cx v Dx)

Now this last sentence is true iff there is something that is either not a cat or is a dog.  But that’s true! 

Only

Consider the sentence

Only mammals are dogs.

It seems to be true.  How should it be symbolized?  An existential would be true even if some dogs are not mammals.

Ex(Mx /\ Dx)

We need to use a universal.  Which of the following should we use?

Vx(Mx -> Dx)            “All mammals are dogs”

Vx(Dx -> Mx)            “All dogs are mammals”

The first is false, and the second is true.  So the second seems to be the way to go.  Generalizing:

Only Fs are Gs                                                 

Vx(Gx -> Fx)

Notice that ‘only’ introduces the predicate that will appear in the consequent of your

symbolization.  This is like ‘only if’ and conditionals.

All Fs and Gs

The rules of thumb for basic categorical sentences usually give one the right symbolization, but they can lead one astray in certain cases.  Consider:

All juniors and seniors are upperclassmen.

This seems to be a universal.  But the following symbolization is problematic.

Vx([Jx & Sx] -> Ux).

It is problematic because nothing is both a junior and a senior.

We could instead use a conjunction of two universals.

Vx(Jx -> Ux)  /\ Vx(Sx -> Ux)

Or we could use a single universal with disjunctive antecedent.

Vx([Jx v Sx] -> Ux).

Or we could use a single universalization of a conjunction.

Vx([Jx -> Ux] /\ [Sx -> Ux])

Both

A similar issue arises with ‘both’.  Consider the sentence:

Both juniors and seniors are upper-classmen.

This seems to be a universal.  We might be tempted to symbolize with

Vx([Jx /\ Sx] -> Ux)

But this is wrong, because nothing is both a junior and a senior.  We could use two universals.

Vx(Jx  -> Ux) /\ Vx(Sx -> Ux).

Or we could use a single universal with a disjunctive antecedent.

Vx([Jx v Sx] -> Ux)

Vacuous Universal Generalizations

Consider the following sentence.

Every purple cow moos.

Is this sentence true or false?  It may be hard to say.  However, our symbolization of this sentence is definitely true.

Vx([Px /\ Cx] -> Mx)

To figure out whether this sentence is true, first drop the universal quantifier phrase.  Then ask whether the result would be true no matter what object is assigned to some object in the domain.  Since there are no purple cows, the antecedent is false no matter what is assigned to.  Therefore, the whole conditional is true.  Therefore, the universal is true.

Think of it this way: all purple cows moo, all none of them. When a sentence of the form

Vx(Fx -> Gx)

We say that it is vacuously true.