Lecture 15: Validity and Predicate
Logic
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Goals Today
✤
Learn the definition of valid and invalid arguments in terms of the
semantics of predicate logic, and look at several examples.
✤
Learn how to get equivalents in predicate logic, and look at how
equivalents can help us show things about validities.
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Recall: Propositional Validity
✤
✤
Here’s the definition we learned in Lecture 12 for formulas of
propositional logic.
We say that ϕ1, …, ϕn ⊨ ψ
if any truth-table which contains columns for each of ϕ1, …, ϕn, ψ has this
feature: whenever a row contains a T in each of the ϕ1, …, ϕn columns, this
row also has a T in the ψ column.
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Defn: Validity in Predicate Logic
✤
Suppose ϕ1, …, ϕn , ψ, are formulas of predicate logic. Then we say
ϕ1, …, ϕn ⊨ ψ if any model M which interprets the language of ϕ1, …, ϕn , ψ has this
feature: if VM(ϕ1)=T, . . . , VM(ϕn)=T then VM(ψ)=T.
✤
Again, we call ϕ1, …, ϕn the premises and we call ψ the conclusion. And
the argument with premises ϕ1, …, ϕn and conclusion ψ is said to be
valid if ϕ1, …, ϕn ⊨ ψ. We may also continue to use the “follows from”
and “has a consequence” locutions.
✤
Again, the way to see the relationship between propositional logic
and predicate logic is as follows: individual models correspond to
individual rows of the truth-table.
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Example 1
✤
Let’s show that ∀x Fx ⊨ Fa.
✤
So we need to show that any
model M in the language with
an individual constant a and
predicate letter F is such that if VM(∀x Fx)=T then VM(Fa)=T
✤
Well, suppose that M is such a
model, and suppose that it’s the
case that VM(∀x Fx)=T. We have
to show that VM(Fa)=T.
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✤
But, by the clause for ∀ in the
semantics, VM(∀x Fx)=T implies
that for all individual constants
c, one has that VM(Fc)=T
✤
In particular, for the individual
constant a, we have that
VM(Fa)=T which is what we wanted to
show.
Example 2
✤
✤
✤
Let’s show that
(∀x Ax) ∨ (∀x Bx) ⊨ ∀x (Ax∨Bx)
✤
Suppose first that VM(∀x Ax)=T.
Then the clause for ∀ says that
VM(Ac)=T for each c. Then the
clause for ∨ implies that
VM(Ac ∨ Bc)=T for each c. But
then the clause for ∀ says that
VM(∀x (Ax∨Bx)) = T.
✤
We argue similarly in the case
where VM(∀x Bx)=T. Hence, we
conclude that in either case we
have VM(∀x (Ax∨Bx)) = T, which
is what we wanted to show.
So suppose M is a model with VM((∀x Ax) ∨ (∀x Bx))=T
We must show we also have
VM(∀x (Ax∨Bx)) = T
Since VM((∀x Ax) ∨ (∀x Bx))=T,
the clause for ∨ says that VM(∀x Ax)=T or VM(∀x Bx)=T
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Definition of Invalidity
✤
Suppose ϕ1, …, ϕn , ψ, are formulas of predicate logic.
✤
Then we say
ϕ1, …, ϕn ⊭ ψ if at least one model M interpreting the language of ϕ1, …, ϕn , ψ has
this feature: VM(ϕ1)=T, . . . , VM(ϕn)=T and VM(ψ)=F.
✤
Again, we call ϕ1, …, ϕn the premises and we call ψ the conclusion. And
the argument with premises ϕ1, …, ϕn and conclusion ψ is said to be
invalid if ϕ1, …, ϕn ⊭ ψ.
✤
There’s an obvious sense in which showing invalidity is easier than
showing validity: to show invalidity, you just need to find one model.
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Example 3
✤
✤
Let’s show that ∀x (Ax∨Bx) ⊭ (∀x Ax) ∨ (∀x Bx)
✤
So we just need to find one
model M with the property that VM(∀x (Ax∨Bx))=T and
VM((∀x Ax) ∨ (∀x Bx))=F
✤
✤
Well, consider our old friend M
with domain natural numbers
and with A interpreted as the
evens and B interpreted as the
odds.
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Let’s check VM(∀x (Ax∨Bx))=T.
By the clause for ∀, we have
that it suffices to show that
VM(Ac∨Bc)=T for all c from M.
But every number is even or
odd, and so we know this!
Let’s check that we indeed have VM((∀x Ax) ∨ (∀x Bx))=F
Well, VM(∀x Ax)=F, since 3 is
not even, and VM(∀x Bx)=F,
since 2 is not odd. So by clause
for ∨, VM((∀x Ax) ∨ (∀x Bx))=F.
Example 3 (Venn Diagrams)
✤
Let’s show that ∀x (Ax∨Bx) ⊭ (∀x Ax) ∨ (∀x Bx)
✤
So we just need to find one
model M with the property that VM(∀x (Ax∨Bx))=T and
VM((∀x Ax) ∨ (∀x Bx))=F
✤
Well, consider our old friend M
with domain natural numbers
and with A interpreted as the
evens and B interpreted as the
odds.
✤
There’s a traditional way of
illustrating these models with
pictures called Venn diagrams:
A
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M
B
More Venn-Diagram Examples
✤
Most simple relationships
between two predicates A,B can
be easily visualized in terms of
Venn Diagram.
✤
Here’s an example of a diagram
of model M where the
following is true:
∀ x (Ax→ Bx)
✤
B
So in terms of this picture, this
sentence represents the
containment of A in B.
A
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M
More Venn-Diagram Examples
✤
Most simple relationships
between two predicates A,B can
be easily visualized in terms of
Venn Diagram.
✤
Here’s an example of a diagram
of model M where the
following is true:
∃ x (Ax ∧ Bx)
✤
B
A
M
So in terms of this picture, this
sentence represents that A and
B overlap.
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More Venn-Diagram Examples
✤
✤
✤
So what would a diagram of ∀ x Ax look like?
Well, it just corresponds to the
claim that everything in the
domain is in A.
A
M
It’s easy to visualize with the
following animation.
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More Venn-Diagram Examples
✤
✤
So what would a diagram of ∃ x Ax look like?
Well, it just corresponds to the
claim that something is in A. So
it corresponds to A just being
any shape within M.
M
A
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Example 4
✤
Let’s show that we have
∀x∃y Rxy ⊭ ∃y ∀x Rxy
✤
Consider the model M, where
Rab iff there’s an arrow from a
to b in the picture:
1
2
3
4
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✤
Let’s check VM(∀x∃y Rxy)=T.
We just need to check that for
each number, there’s an arrow
from that number to another
number. But we see it’s so!
✤
Let’s check VM(∃y ∀x Rxy)=F.
Suppose it was true, that is,
suppose VM(∃y ∀x Rxy)=T. By
the clause for ∃, we would have
VM(∀x Rxc)=T for some c. But c
can’t be 1 (since . . .), and it
can’t be 2 (since . . ), etc.
Ω
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