Lecture 11: Examples of
Semantics of Predicate Logic
1
Outline
✤
I. Venn Diagram Examples
✤
II. Arrow Examples
✤
III. Number System Examples
2
Outline
✤
I. Venn Diagram Examples
✤
II. Arrow Examples
✤
III. Number System Examples
3
Venn Diagrams: Example 1
✤
One way to describe models is with
Venn diagrams.
✤
The domain of the model M is the set of
points inside the box. The interpretation
of the letter A is the auburn circle and
the interpretation of the letter B is the
blue triangle.
✤
This model is a model of VM(∀ x (Ax ➝ Bx))=T
✤
For, suppose p is a name for a point in
M. Then V(Ap ➝ Bp) = T. For, suppose
this was wrong. Then we would have
V(Ap)=T and V(Bp)=F. But then by
clauses for atomics, I(p)∈I(A)=auburn
circle, but not I(p)∈I(B)= blue triangle.
But the auburn circle is contained in the
blue triangle, as we can see from
diagram.
M
I(B)
I(A)
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Venn Diagrams: Moral 1
✤
✤
✤
Recall from the primer on sets that we
say that I ⊆ I’ iff every member of I is
a member of I’.
✤
When we’re drawing sets as shapes
within the Euclidean plane, I ⊆ I’ iff
the shape I is contained in shape I’.
Suppose we now work in a model M
where we have unary predicates A, B,
and where we interpret these as
shapes I(A), I(B) within the Euclidean
plane.
Then the following are equivalent:
✤
1. VM(∀ x (Ax ➝ Bx))=T
✤
2. I(A) ⊆ I(B)
✤
3. The shape I(A) is contained in the
shape I(B)
M
I(B)
I(A)
5
Venn Diagrams: Example 2
✤
The domain of the model M is the set
of points inside the box. The
interpretation of the letter A is the
auburn diamond and the
interpretation of the letter B is the
blue square.
✤
This model is a model of VM(∃ x (Ax ∧ Bx))=T
✤
Let p be a name for one of those points
so that both I(p) ∈ I(A) as well as I(p)
∈ I(B). Then by clauses for atomics,
we have that VM(Ap)=T and
VM(Bp)=T. Then by clause for
conjunction we have VM(Ap ∧ Bp)=T.
Then by clause for existential, we
have that VM(∃ x (Ax ∧ Bx)) = T.
M
✤
I(B)
For, as we can see from the diagram,
there are some points in both the
auburn diamond and the blue square.
I(A)
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Venn Diagram: Moral 2
✤
✤
✤
Recall from primer on sets that ∅ is our
abbreviation for the empty set, and that
I∩I’ is our abbreviation for the
intersection of two sets, i.e. what they
have in common.
✤
If I, I’ are shapes in the Euclidean plane,
one has that I∩I’ ≠ ∅ iff they have some
common area of overlap.
Then the following are equivalent:
✤
1.
✤
2.
✤
3.
VM(∃ x (Ax ∧ Bx))=T
I(A)∩I(B) ≠ ∅
The shapes I(A), I(B) overlap.
M
Suppose we now work in a model M
where we have unary predicates A, B,
and where we interpret these as shapes
I(A), I(B) within the Euclidean plane.
I(B)
I(A)
7
Venn Diagrams: Example 3
✤
The domain of the model M is the set
of points inside the box. The
interpretation of the letter A is the
auburn rectangle and the
interpretation of the letter B is the
blue rectangle. The idea is that they
cover all of the model M.
✤
This model is a model of VM(∀ x (Ax ∨ Bx))=T
✤
✤
Then I(p)∈I(A) or I(p)∈ I(B). By clause
for atomics, VM(Ap)=T or VM(Bp)=T.
Then by clause for disjunction, we
have that VM(Ap ∨ Bp)=T. We can do
the same for any other point. So by
clause for universal quantifier, we
have that VM(∀ x (Ax ∨ Bx))=T.
M
I(A)
For, suppose p is a name for a point in
the domain. Then it’s contained in the
auburn rectangle or blue rectangle.
8
I(B)
Venn Diagrams: Moral 3
✤
✤
Recall from our primer on sets that I∪I’ is the union of the two sets I, I’, i.e.
what happens when you put the two
sets together. If I, I’ are shapes in the
Euclidean plane, then I∪I’ is what you
get by gluing the two shapes together.
✤
Suppose we now work in a model M
where we have unary predicates A, B,
and where we interpret these as
shapes I(A), I(B) within the Euclidean
plane.
Then the following are equivalent:
✤
1. VM(∀ x (Ax ∨ Bx))=T
✤
2. I(A)∪I(B) = M
✤
3. When you glue the shape I(A)
together with the shape I(B), you get
all of M.
M
I(A)
9
I(B)
Outline
✤
I. Venn Diagram Examples
✤
II. Arrow Examples
✤
III. Number System Examples
10
Arrow Models: Example 1
✤
Suppose that the domain of our
model M consists of some small
finite collection of numbers. In the
example displayed, it is four
numbers 1,2,3,4. Assume that we
have constants c1, c2, c3, c4 in our
language such that I(c1)=1, I(c2)=2, I(c3)=3, I(c4)=4.
✤
We further assume that the binary
relation R is interpreted so that:
✤
✤
Then one can verify in our model
that VM(∃ y ∀ x Rxy)=T. For, since
there is an arrow from m to 4 for
each m, we have VM(Rcmc4)=T. Then
by clause for universals, VM(∀ x Rxc4)=T. Then by clause for existentials, VM(∃ y ∀ x Rxy)=T.
1
4
M
Rmn is true in the model iff there
is arrow going from m to n.
2
3
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Arrow Models: Moral 1
✤
✤
Suppose that the domain of our
model M consists of some small
finite collection of numbers.
✤
Assume that the binary relation R
is interpreted so that:
✤
Then the following are equivalent:
✤
1. VM(∃ y ∀ x Rxy)=T
✤
2. There’s a point such that
every point has an arrow to it.
Rmn is true in the model iff there
is arrow going from m to n.
1
4
M
2
3
12
Arrow Models: Example 2
✤
Consider the arrow model
displayed in the lower right.
✤
We have
✤
VM(Rc1c2)=T, and so VM(∃ y Rc1y)=T
✤
VM(Rc2c3)=T, and so VM(∃ y Rc2y)=T
✤
VM(Rc3c4)=T, and so VM(∃ y Rc3y)=T
✤
VM(Rc4c4)=T, and so VM(∃ y Rc4y)=T
✤
Since our model only has four
points, it follows from the clause
for the universal quantifier that
VM(∀ x ∃ y Rxy)=T
1
4
M
2
3
13
Arrow Models: Moral 2
✤
✤
Suppose that the domain of our
model M consists of some small
finite collection of numbers.
✤
Assume that the binary relation R
is interpreted so that:
✤
Then the following are equivalent:
✤
1. VM(∀ x ∃ y Rxy)=T
✤
2. Every point has an arrow to
some point (perhaps itself).
Rmn is true in the model iff there
is arrow going from m to n.
1
4
M
2
3
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Outline
✤
I. Venn Diagram Examples
✤
II. Arrow Examples
✤
III. Number System Examples
15
Number Systems Examples
✤
These examples are more
difficult, and have less of a clear
“moral.”
✤
This is one kind of evidence
that one can amass for the
correctness of a semantics.
✤
The reason for including them
is: we all already have a lot of
beliefs about the basic number
systems, about what is true and
what is false on them.
✤
✤
We show in the next slides that
our formal semantics predicts
these ordinary beliefs.
As we’ll see later, for other
phenomena (like definite
descriptions) there are
competing semantics, and so
predicting the phenomena is
one way that one might
adjudicate between competing
semantics.
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Number System Examples 1
✤
✤
✤
Consider a language which has
infinitely many constant
symbols c0, c1, c2, c3, . . . . . and
one binary relation symbol ≤.
One model for this is given by
domain ℕ = {0,1,2,3,4,. . . } and
with I(c0)= 0, I(c1)= 1, I(c2)= 2,
etc. and I(≤) = {〈x,y〉∈ℕ: x≤y}
Consider the sentence:
∀x ∃ y x≤y
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✤
Example: Vℕ(∀x ∃ y x≤y)=T.
✤
By the clause for universal
quantifiers, we must show that Vℕ(∃ y cn≤y)=T for all n.
✤
Fix n. By the clause for the
existential quantifier, we must
show that Vℕ(cn≤d)=T for at
least one constant d.
✤
But just choose d=cn.
Number System Examples 2
✤
✤
✤
Consider a language which has
infinitely many constant
symbols c0, c1, c2, c3, . . . . . and
one binary relation symbol ≤.
One model for this is given by
domain ℕ = {0,1,2,3,4,. . . } and
with I(c0)= 0, I(c1)= 1, I(c2)= 2,
etc. and I(≤) = {〈x,y〉∈ℕ: x≤y}
Consider the sentence:
∃ x ∀ y x≤y
18
✤
Example: Vℕ(∃ x ∀ y x≤y)=T.
✤
By the clause for existential
quantifiers, we must show that Vℕ(∀ y c≤y)=T for at least one
constant c from L.
✤
So consider c=c0. Then we have
Vℕ(∀ y c0≤y)=T by the clause for
the universal quantifier and
since Vℕ(c0≤d)=T for all
constants d from L.
Number System Examples 3
✤
✤
✤
Consider a language which has
infinitely many constant symbols
c0, c1, c2, c3, . . . . . and one binary
relation symbol ≤.
One model for this is given by ℤ = {. . . ., -4,-3,-2,-1,0,1,2,3,4,. . . }
and I(c0) = 0, I(c1)=1, I(c2)=-1, etc.
and I(≤) = {〈x,y〉∈ ℤ: x≤y}
Consider the sentence:
∀x ∃ y x≤y
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✤
Example: Vℤ(∃ x ∀ y x≤y)=F.
✤
For, suppose not. Then we would
have Vℤ(∃ x ∀ y x≤y)=T. Then we
would have:
(*)
Vℤ(∀ y d≤y)=T for some constant d from L.
✤
But consider a constant c such that
I(c) = I(d)-1. By (*) and clause for
universal quantifier, (**)
Vℤ(d≤c)=T
Then I(d)≤I(c) =I(d)-1. But
subtracting makes numbers smaller
Ω
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