Lecture 16: Validity & Equivalences
1
Outline
✤
I. Transferring Equivalences from Propositional to Predicate Logic
✤
II. Equivalent Formulations of Validity and Deduction Theorem
2
Outline
✤
I. Transferring Equivalences from Propositional to Predicate Logic
✤
II. Equivalent Formulations of Validity and Deduction Theorem
3
Transferring Valid Arguments
✤
So it turns out that we can
transfer valid arguments in
propositional logic into valid
arguments in predicate logic.
✤
For instance, we showed that
the following were valid
arguments in propositional
logic:
p, p→q ⊨ q
¬p, p∨q ⊨ q
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✤
It turns out that if ϕ, ψ are
sentences of predicate logic,
then we have that the following
are valid arguments in
predicate logic:
ϕ, ϕ → ψ ⊨ ψ
¬ϕ, ϕ∨ ψ ⊨ ψ
✤
This is because the clauses for
¬,∧,∨,➝ are identical in
propositional and predicate
logic.
Transferring Invalidities
✤
✤
✤
Some care has to be exercised in
translating invalid arguments
of propositional logic into
predicate logic.
What’s true is that if you have
an invalidity of propositional
logic, then some substitution
instance will be an invalidity of
predicate logic.
However, not all substitution
instances will be invalid.
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✤
Consider the invalid argument
q, p→q ⊭ p
✤
It turns out that the following
substitution instance is in fact
valid:
∀x Fx, (Fc→∀x Fx) ⊨ Fc
✤
Intuitively, the reason is
because when you substitute,
what you replace p and q with
can exhibit new dependencies
on one another.
Transferring
Equivalences
✤
✤
Equivalences between ϕ and ψ
are just a special case of
validities, namely in which we
have both ϕ ⊨ ψ and ψ ⊨ ϕ
name of
equivalence:
this is equivalent to this over here:
DeMorgan Law
¬(ϕ ∨ ψ)
(¬ ϕ ∧ ¬ ψ)
DeMorgan Law
¬(ϕ ∧ ψ)
(¬ ϕ ∨ ¬ ψ)
Contraposition
(ϕ➝ ¬ψ)
(ψ ➝ ¬ϕ)
Distribution
(ϕ ∧ (ψ ∨ χ))
((ϕ ∧ ψ) ∨ (ϕ ∧ χ))
Distribution
(ϕ ∨ (ψ ∧ χ))
((ϕ ∨ ψ) ∧ (ϕ ∨ χ))
Hence, all of the equivalents
that you memorized for
propositional logic also become
equivalents of predicate logic.
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Helpful Equivalents from Lec 11
(cf. Gamut p.100)
Sentence
Equivalent Sentence
1
¬ (∀x ϕ(x))
∃x ¬ϕ(x)
2
¬ (∃x ϕ(x))
∀x ¬ϕ(x)
3
∀x (ϕ(x) ∧ ψ(x))
(∀x ϕ(x)) ∧ (∀x ψ(x))
4
∃x (ϕ(x) ∨ ψ(x))
(∃x ϕ(x)) ∨ (∃x ψ(x))
5
∀x ∀ y Rxy
∀y ∀ x Rxy
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∃ x ∃ y Rxy
∃y ∃x Rxy
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Example
Let’s show that
✤
✤
We know from our list of
equivalents that ¬∀x Fx is
equivalent to ∃x ¬Fx. Hence we
know that ((¬∀x Fx) ∨ (¬∀x Gx))
is equivalent to
((∃x ¬Fx) ∨ (∃x ¬Gx))
✤
So we not only just showed that
¬(∀x Fx ∧ ∀x Gx) ⊨(∃x¬Fx) ∨ (∃x ¬Gx)
We can do this without even
thinking about models.
✤
We know by DeMorgan that ✤
¬(∀x Fx ∧ ∀x Gx)
is equivalent to ¬(∀x Fx ∧ ∀x Gx) ⊨(∃x¬Fx) ∨ (∃x ¬Gx)
But we also showed that (∃x¬Fx) ∨ (∃x ¬Gx) ⊨ ¬(∀x Fx ∧ ∀x Gx)
((¬∀x Fx) ∨ (¬∀x Gx))
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Outline
✤
I. Transferring Equivalences from Propositional to Predicate Logic
✤
II. Equivalent Formulations of Validity and Deduction Theorem
9
Recall: Definitions of Validity and
Tautology
✤
So let’s work in propositional logic for the moment, since it’s simpler.
✤
Recall in this setting that we define:
✤
We say that the argument with premises ϕ1, …, ϕn and conclusion ψ is
valid (written: 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.
✤
We also say that ψ is a tautology if when you draw the truth-table for ψ, it
is the case that every row has a ’T’ in the column for ψ.
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Equivalent Formulations of Validity
(cf. Gamut p. 123 Theorem 2)
✤
Theorem. The following are equivalent:
✤
1. ϕ1, …, ϕn ⊨ ψ
✤
2. ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ) is a tautology.
✤
On the next slides, we’ll prove this result. But now just a brief remark
on what it means. It means that instead of showing a validity, you can
show instead that something is a tautology.
✤
It should resonate with the feeling that you might have had that
there’s something very similar between establishing that something is
a validity and the truth-table for the → symbol.
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Proof of Result 1/2
✤
✤
So suppose that ϕ1, …, ϕn ⊨ ψ.
Now let’s show that the
following is a tautology:
((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ)
Since ϕ1, …, ϕn ⊨ ψ, if we made
the truth-table which had
columns for all of ϕ1, …, ϕn, ψ,
then for any row which had T’s
in all the ϕ1, …, ϕn columns, we
would have a T in the ψ
column.
ϕ1 ϕ2
⋅⋅⋅
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✤
So we never have a row where
ϕ1, …, ϕn are all T and ψ is F.
✤
So then expand this table by
columns for both (ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn)
as well as ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ).
✤
If we fill in the last column
we’re never going to get an F.
ϕn
ψ
(ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn)
(ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn)→ ψ
Proof of Result 2/2
✤
Suppose that the following is a
tautology: ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ).
Let’s show that ϕ1, …, ϕn ⊨ ψ.
✤
Since ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ) is a
tautology, when we draw it’s
truth-table we always get T’s in
its column. But this table
already contains columns for
the formulas ϕ1, …, ϕn, ψ.
ϕ1
Suppose that ϕ1, …, ϕn ⊭ ψ.
Then there would be a row
where ϕ1, …, ϕn all were T, and
yet ψ received an F. But then
(ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) would have T at
that row. But since ψ has an F,
we would have an F in the
column for ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ).
✤
ϕ2
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⋅⋅⋅
ϕn
ψ
(ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn)
(ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn)→ ψ
Equivalent Formulations of Validity
(cf. Gamut p. 123 Theorem 2)
✤
So we just proved this theorem. This kind of suggests that the
consequence relation ⊨ acts a lot like the arrow. Our next theorem also
illustrates this idea.
✤
Theorem. The following are equivalent:
✤
1. ϕ1, …, ϕn ⊨ ψ
✤
2. ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ) is a tautology.
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The Deduction Theorem (cf.
Gamut p. 122 Theorem 1)
✤
Deduction Theorem. The following are equivalent:
✤
1. ϕ1, …, ϕn, ϕ ⊨ ψ
✤
2. ϕ1, …, ϕn ⊨ ϕ→ψ
✤
We’ll prove this theorem on the next slides. But what it says is that
you can move premises from left to right across the consequence
relation so long as you put them in front of an arrow.
✤
Before we get to the proof of the deduction theorem, let’s note on the
next slide a simple and illustrative application of the deduction
theorem.
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Simple Application of the
Deduction Theorem
✤
✤
Here’s a simple application of
the deduction theorem.
Let’s use it to show that (i)
ψ ⊨ ϕ → (ϕ → ψ)
✤
By the deduction theorem, this
is equivalent to (ii)
ψ, ϕ ⊨ (ϕ → ψ)
✤
By the deduction theorem
again, this is equivalent to (iii) ψ, ϕ, ϕ ⊨ ψ
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✤
But it’s not hard to see, by
recourse to the definition of
validity, that (iii) is true.
✤
For, if we have a truth-table
which contains ψ and ϕ, we
then of course any row in
which all three of ψ, ϕ, ϕ have
T’s is going to be a row in
which ϕ is true.
Proof of the Deduction Theorem, 1/2
✤
So suppose ϕ1,…, ϕn, ϕ ⊨ ψ. We
show that ϕ1, …, ϕn ⊨ ϕ→ψ.
✤
So suppose you have a truthtable with columns for ϕ1,…,
ϕn, ϕ, ψ. Since ϕ1,…, ϕn, ϕ ⊨ ψ,
for any row where all of ϕ1,…, ϕn, ϕ have T’s, it is also
the case that ψ is a T.
✤
Then add onto that truth-table
a column for ϕ→ψ.
ϕ1
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✤
Suppose ϕ1, …, ϕn ⊭ ϕ→ψ.
Then there would be a row of
this table where ϕ1, …, ϕn was
T and where ϕ→ψ was F. Then
this row must have ϕ be T and
ψ be F.
✤
But then we contradict our
assumption ϕ1,…, ϕn, ϕ ⊨ ψ.
ϕ2
⋅⋅⋅
ϕn
ϕ
ψ
ϕ→ψ
Proof of the Deduction Theorem, 2/2
✤
So suppose ϕ1, …, ϕn ⊨ ϕ→ψ.
We show ϕ1,…, ϕn, ϕ ⊨ ψ.
✤
So suppose you have a truthtable with columns for ϕ1,…,
ϕn, ϕ→ ψ. Since ϕ1, …, ϕn ⊨
ϕ→ψ, for any row where all of ϕ1,…, ϕn have T’s, it is also the
case that ϕ→ ψ is a T.
✤
Then this table already has
columns for ϕ, ψ.
ϕ1
✤
Suppose ϕ1,…, ϕn, ϕ ⊭ ψ. Then
there would be a row of this
table where ϕ1,…, ϕn, ϕ was T
and where ψ was F. Then we
would have ϕ→ ψ be F.
✤
But then we contradict our
assumption ϕ1, …, ϕn ⊨ ϕ→ψ.
ϕ2
18
⋅⋅⋅
ϕn
ϕ
ψ
ϕ→ψ
Recap: The Two Results
✤
✤
Theorem. The following are equivalent:
✤
1. ϕ1, …, ϕn ⊨ ψ
✤
2. ((ϕ1 ∧ ⋅ ⋅ ⋅ ∧ ϕn) → ψ) is a tautology.
Deduction Theorem. The following are equivalent:
✤
1. ϕ1, …, ϕn, ϕ ⊨ ψ
✤
2. ϕ1, …, ϕn ⊨ ϕ→ψ
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✤
These two results are significant
because they suggest that even though
the consequence relation ⊨ cannot be
put into the premises and conclusion,
nevertheless, it acts a lot like the arrow.
✤
We’ve proven these results for
propositional logic, since the proofs are
more intuitive in that setting. But the
theorems hold also for predicate logic.
Ω
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