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FOL Symbolization 3: Mixed Domains,
Negation, and Identity
These Practice Problems use examples which try to bring out (a) some of the difficulties in
rendering fairly ordinary English claims in quantifier language, and (b) emphasise some of the
things you can do once the identity predicate is combined with quantifiers and negation.
The way to approach these problems is as follows. Read the relevant parts of the Slides and
Textbook, and take notes as you normally would. Then look at a group of practice examples
below, and try to do these without consulting the Slides/Textbook. In many cases, you're given
strong guidance before (sometimes after) the question. If Carnap accepts the answer, good. If
it doesn't, then try to see what's wrong in it, and try various other things. It may be the
brackets, which are important here, among other things, for determining how the variables
(think of them here as pronouns!) tie to their quantifiers. As a last resort, look back at the
Slides, but do so in a pointed way: the kind of reaction you want in yourself is 'Oh, that's how
you do it, it's really the existential quantifier that ...', so you see how to do it, and realise why
you do it that way. In many cases, you'll be able to get the right formulations by reading them
off the Slides or from the Textbook. But try them yourself first, and then if you do resort to
these written sources, then enter the example, try to see why it's correct, internalise the
correct formulation, refresh the page to erase the answer, and try again to recapture it.
As before, to enter logical symbols with the keyboard, use:
not ¬
- , ~
and ∧
/\ , &
or ∨
\/
if then →
-> , >
if and only if ↔︎
<-> , <>
contradiction ⊥
!? , _|_
universal quantifier ∀ A , @
existential quantifier ∃ E , 3
We'll use the following key from the Slides:
Domain: people
H(x) x is a hero
g Greta
V(x) x is a villain
a Autumn
C(x) x wears a cape
I(x)
x inspires
A(x,y) x admires y
Y(x,y) x is younger than y
The indefinite article 'a', 'an'
The indefinite article sometimes works like an existential determiner, and sometimes like a
universal.
'Greta admires a hero' surely says that there is a hero whom Greta admires, so it can be
symbolized the same way as 'Some heroes are admired by Greta'.
FOL.3.1
Greta admires a hero.
Greta admires a hero.
But note that 'If someone is a hero, Greta admires them' doesn't say that. Rather, it says the
same as 'All heroes are admired by Greta'.
FOL.3.2
If someone is a hero, Greta admires them.
If someone is a hero, Greta admires them.
'Anything', 'any'
'Anything', 'anyone', and 'any' are determiners that sometimes call for an existential and
sometimes a universal quantifier.
Compare the following:
FOL.3.3
Any hero admires Greta.
Any hero admires Greta.
FOL.3.4
If any hero admires Greta, Autumn does.
If any hero admires Greta, Autumn does.
Mixed quantifiers
If the sentence you have to symbolize contains more than one determiner, you have to proceed
in steps. E.g., 'All heroes are younger than a villain' says that every hero has the property of
being younger than a villain. So your schema will start '∀x(V(x) → x younger than a villain'. And
the property of being younger than a villain can be expressed by '∃y(V(y) ∧ Y(x,y)' (i.e., x is
younger than a villain iff there is some villain that x is younger than).
FOL.3.5
Ax(H(x)!"3y(Y(x,y)&V(y)))
All heroes are younger than a villain.
✓
'Some hero who is younger than a villain wears a cape', on the other hand, says that someone
is a hero, satisfies 'x is younger than a villain', and wears a cape. Thus it would have the partial
form: '∃x(H(x)∧x is younger than a villain ∧C(x))'.
FOL.3.6
Ex((H(x)!#3y(Y(x,y)&V(y))!#C(x)))
Some hero who is younger than a villain wears a cape
✓
Mixed domains
You'll have noticed that 'x wears a cape' contains a determiner phrase, 'a cape'. So if our
domain contains capes as well as people, and our symbolization key contains E(x) for 'x is a
cape' and R(x,y) for 'x wears y' we can express the property 'x wears a cape' by
'∃y(E(y)∧R(x,y))'. However, to symbolize 'everyone' and 'someone' we now also requires a
predicate for 'x is a person'.
Domain: people, items of clothing
H(x) x is a hero
g Greta
V(x) x is a villain
a Autumn
P(x) x is a person
E(x) x is a cape
I(x)
x inspires
A(x,y) x admires y
Y(x,y) x is younger than y
R(x,y) x wears y
B(x,y) x belongs to y
FOL.3.7
Ax(P(x)!"3y(E(y)&R(x,y)))
Everyone wears a cape
✓
Donkey sentences
Some sentences in which a pronoun refers to an embedded determiner phrase are hard to
symbolize. 'Every hero who admires someone is younger than they are' is an example: 'they'
refers to 'someone', but it's the 'someone' that's misleading here. Sentences of this kind are
called 'donkey sentences' because the medieval logician Walter Burley first brought attention
to them with the example 'Every farmer who owns a donkey sees it'. This is another example
where the 'a' and the 'someone' have to be dealt with with universal quantifiers and not
existential, which is what the English ('someone, 'a particular donkey') at first suggests. But the
sentence is surely talking about any farmer and then any donkey owned by that farmer, not just
some donkey the farmer owns. Another way to think of it is this: the sentence will be false just
in case there is some farmer and some donkey owned by that farmer that the farmer doesn't
see. Therefore it will be true just in case that is negated, and this will lead us to 'For all x, and all
y (If x is a farmer and y is a donkey and x owns y, then x sees y)'. Alternatively: 'For all x (If x is a
farmer then for all y(y is a donkey and x owns y, then x sees y))'.
Now go back to 'Every hero who admires someone is younger than they are', and remember to
restrict the 'everyone' determiner to people!
FOL.3.8
Ax[(H(x)&P(x))!"Ay((P(y)&A(x,y))!"Y(x,y))]
Every hero who admires someone is younger than they are.
✓
Identity
Different variables don't require different objects. So (in a domain consisting only of people),
∀x∀y A(x,y) means that everyone admires everyone, not that everyone admires everyone else.
Similarly, ∃x∃y A(x,y) doesn't mean that someone admires someone else: it is also true if
someone admires themselves. However, the identity predicate can be used to supply the 'else'
part: ∀x∀y(¬x=y→A(x,y)) is true if everyone satisfies 'x admires everyone who is not x'. Let's try
this out with 'Someone admires everyone else', but now also use 'P(x)' to restrict 'everyone'
and 'someone' to people (i.e., say that every person admires every person different from them).
FOL.3.9
3x[P(x)&∀y((P(y)&~x=y)!"A(x,y))]
Someone admires everyone else.
✓
Uniqueness, singular only
To say that someone is a hero is to say that at least one person satisfies 'H(x)', and so can be
symbolized by '∃x H(x)'. To say that there is exactly one hero is to say that there is a person
who satisfies 'H(x)', and no-one else is a hero. To say 'No-one other than x is a hero' is to say
that no hero exists who is different from x.
FOL.3.10
3x[H(x)&∀y(H(y)!"x=y)]
There is exactly one hero.
✓
The same idea can be used to symbolize sentences containing singular 'only'. We already know
how to symbolize plural 'only' as in 'only heroes'. But what about 'Only Greta is a hero'? That
says that Greta is a hero, and also no-one other than Greta is a hero.
FOL.3.11
∀x(H(x)!!%x=g)
Only Greta is a hero.
✓
'At least', 'at most', 'exactly'
Since different variables don't require different objects, '∃x∃y(H(x)∧H(y))' does not say that
there are at least two heroes. It just says that someone is a hero, and someone is a hero, which
happens to be true if actually only one person is a hero. If you want x and y to refer to different
things, you have to include '¬x = y' explicitly.
For the rest of the questions, we will assume for simplicity that we again referring to a domain
which just has people in it, so we don't have to restrict the quantifiers.
FOL.3.12
3x3y(H(x)&H(y)&~x=y)
There are at least two heroes.
✓
This also lets us say that there is at most one hero: we simply deny that there are at least two.
FOL.3.13
∀x∀y[(H(x)&H(y))!"x=y]
There is at most one hero.
✓
Another way of symbolizing this would be to say that if x and y are heroes, x and y are actually
the same, so there can't be two (or more).
But what about 'There are at most two heroes'? For this, we have to deny that there are at least
three heroes. This in turn requires three ∃x, ∃y, ∃z and we have to say that any two of x, y, z are
different. In other words, '¬x = y ∧ ¬y = z' by itself is not enough.
FOL.3.14
3x3y3z[(H(x)&H(y)&H(z))&(~x=y&~x=z&~y=z)]
There are at least three heroes.
✓
To say that there are at most two heroes, put an ¬ in front of your symbolization of 'There are at
least three heroes'. Alternatively, think of it this way: Of any heroes x, y, z, at least two have to
be the same (i.e., x = y or x = z or y = z).
FOL.3.15
∀x∀y∀z[(H(x)&H(y)&H(z))!"((x=y|x=z)|y=z)]
There are no more than two heroes.
✓
We already know how to say that there is exactly one hero: some x is a hero, and there is no y
that's different from x and also a hero. To say there are exactly two heroes is to say both that
there are at least two and that there are at most two. Alternatively, think of it this way: there are
two different heroes, and any hero is one of them.
FOL.3.16
3x3y[H(x)&H(y)&~x=y&∀z(H(z)!"(x=z|y=z))]
There are exactly two heroes.
✓
'The', 'both', 'neither', singular possessives
The definite article 'the' is used to form determiner phrases called definite descriptions, such
as 'the hero' or 'the youngest villain'. Grammatically, we use definite descriptions in the same
way we use indefinite descriptions ('a hero') or quantifiers ('all heroes', 'no villains'). So 'The
hero is tall' looks like it should be 'Tall(the hero)'. But this is as wrong as representing 'All
heroes are tall' by 'Tall(all heroes)'. So how should we symbolize them?
According to Russell's analysis of definite descriptions, a sentence containing a definite
description, 'The A is B', should be symbolized so that it is true iff (a) exactly one thing is A and
(b) that thing is also B. One way of doing that is to symbolize it as
∃x[(A(x)∧∀y(A(y)→x=y))∧B(x)].
Definite descriptions can be used also to symbolize singular possessives. E.g., 'Autumn wears
Greta's cape' can be paraphrased first as 'Autumn wears the cape that belongs to Greta'. Then
we can use Russell's way of treating definite descriptions.
FOL.3.17
∃x[E(x)&B(x,g)∧∀y((E(y)&B(y,g))→x=y)∧R(a,x)]
Autumn wears Greta's cape.
✓
Russell's idea can also be used to deal with 'both' and 'neither', used not as sentence
connectives (as in 'Neither Autumn nor Greta are villains') but as determiners 'Neither villain is
younger than Greta'. That sentence is true iff (a) there are exactly two villains and (b) each of
them is not younger than Greta.
FOL.3.18
3x3y[V(x)&V(y)&~x=y&∀z(V(z)!"(x=z|y=z))&~Y(x,g)&~Y(y,g)]
Neither villain is younger than Greta.
✓
If we wanted to treat 'Both villains are younger than Greta', we would do exactly the same, but
would just not negate the 'Y(x,y)' predicate.
An Open Tower project. Copyright 2015-2023 G. Leach-Krouse <gleachkr@ksu.edu> and J.
Ehrlich.
This work has been supported by a Munson-Simu Faculty Star award at Kansas State
University.