Symbolic Logic II

Propositional Logic
Axiomatic proofs in propositional logic
We wil now consider a different approach to proof theory, using
axiomatic (or “Hilbert-style”) systems.
Axiomatic proofs are harder to construct than sequent proofs
because we cannot use conditional proofs or reductio ad absurdum.
An axiomatic proof is a series of formulas, the last of which is the
conclusion of the proof. Each line in the proof must be justified in
one of two ways: it may be inferred by a rule of inference from
earlier lines in the proof, or it may be an axiom.
Brandon C. Look: Symbolic Logic II, Lecture 3
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In order to apply the axiomatic method, we need the following:
I
a set of rules, and
I
a set of axioms.
One of our rules is going to be modus ponens:
(φ → ψ)
ψ
φ
MP
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Definition of axiomatic proof from a set: Where Γ is a set of
wffs and φ is a wff, an axiomatic proof from Γ is a finite sequence
of wffs whose last line is φ, in which each line either (i) is an axiom
or (ii) is a member of Γ, or (iii) follows from earlier wffs in the
sequence of a rule.
Definition of an axiomatic proof: An axiomatic proof of φ is an
axiomatic proof of φ from ∅ (i.e., a finite sequence of wffs whose
last line is φ, in which each line either (i) is an axiom, or (ii)
follows from earlier wffs in the sequence via a rule).
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We will write “Γ ` φ” when we mean that φ is provable from the
set of wffs Γ. (For “Γ ` φ” just think “Γ proves φ.” Also, “Γ ` φ”
means “φ is a syntactic consequence of Γ”, whereas “Γ |= φ”
means “φ is a semantic consequence of Γ”.)
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Axiomatic System for PL:
I
Rule: MP
I
Axioms: The result of substituting wffs for φ, ψ, and χ in any
of the following schemas is an axiom:
pφ → (ψ → φ)q
p(φ → (ψ → χ)) → ((φ → ψ) → (φ → χ))q
p(∼ ψ →∼ φ) → ((∼ ψ → φ) → ψ)q
Brandon C. Look: Symbolic Logic II, Lecture 3
(PL1)
(PL2)
(PL3)
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Propositional Logic
Soundness of Propositional Logic
Two very important results of metalogic:
1. Soundness of PL: Every PL-theorem is PL-valid.
2. Completeness of PL: Every PL-valid wff is a PL-theorem.
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Why is this important?
Consider, first, Soundness. Here the claim is often represented as
“if ` φ then |= φ”. And this means that whatever is provable by
propositional logic is valid (or true given true premises). So, you
can’t go from true premises, use our rules of inference and so on,
and arrive at something that is false.
Next, Completeness. Here the claim is represented as “if |= φ then
` φ”. And this means that everything that is true or valid is
provable in propositional logic.
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So, there is no valid consequence that is not derivable from
propositional logic. And our logical syntax matches up with our
semantics (for every semantic consequence in PL is a syntactic
consequence in PL and every syntactic consequence in PL is a
semantic consequence in PL).
That’s a big deal.
Brandon C. Look: Symbolic Logic II, Lecture 3
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Propositional Logic
Aside: method of induction
In order to prove both the soundness and completeness of PL, we
have to discuss an important method of proof: the method of
induction.
Basically, the idea is simple: if we can prove a particular property
of a first object in some group and show that whenever one object
in the group has the property the next one must have the property,
then we can conclude that all objects have that property.
In other words, we need to proceed in two steps:
I
the base case (b)
I
the inductive step (i)
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So, when want to establish that every wff has a certain property, p,
we have to establish the following two claims:
(b) every atomic wff has property p, and
(i) for any wffs φ and ψ, if both φ and ψ have property p, then
the wffs ∼ φ and φ → ψ have property p.
Brandon C. Look: Symbolic Logic II, Lecture 3
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Propositional Logic
The proof of soundness
The basic idea behind this proof is this. We are going to take our
axioms as the starting points or base case(s) and show that they
are PL-valid. Then we are going to show that our inference rule,
modus ponens, preserves validity. If we do this, we can claim that
every theorem in PL is valid.
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So, here’s Sider’s proof:
Base case: We need to show that every PL-axiom is valid. Start
with PL1, pφ → (ψ → φ)q. (PL2 and PL3 will be exercises.)
Suppose for a reductio that some instance of PL1 is invalid; that
means for some PL-interpretation I , VI (φ → (ψ → φ)) = 0.
Thus, VI (φ) = 1 and VI (ψ → φ) = 0. Given the latter,
VI (φ) = 0. Since this leads to a contradiction, PL1 must be valid.
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Induction step: We assume that every line in a proof up to a
certain point is valid, and then we show that, if one adds another
line that follows from the earlier lines by modus ponens, then that
line must be valid, too. So, this means that we are claiming that
some formula ψ can be inferred from two earlier lines containing
the formulas φ → ψ and φ. We must show that ψ is a valid
formula — that VI (ψ) = 1 for every PL-interpretation I . By the
inductive hypothesis, all earlier lines in the proof are valid, and
hence both φ → ψ and φ are valid. Thus, VI (φ) = 1 and
VI (φ → ψ) = 1. But if VI (φ) = 1, then VI (ψ) cannot be 0, for
if it were, then VI (φ → ψ) would be 0, and it is not. Thus,
VI (ψ) = 1.
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Another proof of Soundness:
Suppose Γ ` φ. Now, one metatherem of PL is that Γ ` φ iff there
is a finite subset ∆ of Γ such that ∆ ` φ.
(i) If ∆ is empty, then ` φ and therefore φ is a theorem and Γ ` φ.
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(ii) If ∆ is not empty, let φ1 , . . . , φn be the members of ∆. Then,
φ1 , . . . , φn ` φ. By the Deduction Theorem (which we will discuss
next time), applied as many times as necessary,
` φ1 → (φ2 → (. . . (φn → φ))). A consequence of this is that
|= φ1 → (φ2 → (. . . (φn → φ))). And, by the definition of true for
an interpretation of PL, there is no model of {φ1 , . . . , φn }, that is,
of ∆, that is not also a model of φ. Therefore, since ∆ is a subset
of Γ, there is no model of Γ that is not also a model of φ; that is,
either Γ has no model or every model of Γ is also a model of φ.
That is, Γ |= φ. So, if Γ ` φ, then Γ |= φ.
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