Download Phil 312: Intermediate Logic, Precept 7.

Phil 312: Intermediate Logic, Precept 7.
Alejandro Naranjo Sandoval, [email protected]
1879 Hall, 227
http://scholar.princeton.edu/ansandoval/classes/phi-312-intermediate-logic
1. Boolean Algebras: Examples.
• The main goal of today’s precept will be to familiarize you with the main concepts
of Boolean algebra. Recall: a Boolean algebra B is a set together with a unary
operation ¬, two binary operations ∧ and ∨, and designated elements 0 ∈ B and 1 ∈ B
which satisfy the equations which Prof. Halvorson wrote on the board yesterday. (Note
that perhaps it would be better to refer to these operations by the symbols ¬B , ∧B , ∨B ;
and to the designated elements by 0B , 1B , since these all vary from Boolean algebra.
We don’t do this since it is usually clear from context what we’re talking about.)
• Let’s look at some examples.
i In lecture, we covered the important example of the powerset of a set X (where
X doesn’t need to be finite). Here ∅ = 0B and X = 1B , ∩ is ∧B , ∪ is ∨B , and
X \ A is ¬B A, for any A ∈ P(X).
ii Consider B6 = {1, 2, 3, 6}, where 1 = 0B6 , 6 = 1B6 , and, for any x, y ∈ B6 ,
x ∧ y = gcd(x, y), x ∨ y = lcm(x, y), and ¬x = 6/x. Note then that ¬3 = 2, and
¬2 = 3. (Visualize this algebra as a diamond with 4 vertices, with 6 on top and
1 on the bottom). Try doing some operations in B6 . Does this Boolean algebra
look familiar?
iii Similarly, consider B30 = {1, 2, 3, 5, 6, 10, 15, 30} which operations defined as in B6
(except that ¬x = 30/x) (Visualize this algebra as a cube standing on a vertex.
This bottom vertex is 1. The top one is 30.) Is there a homomorphism between
f : B6 → B30 ?
iv Another very important Boolean algebra: consider B = {[p]R | p ∈ Sent(Σ)}, for
some signature Σ, and where R is the equivalence relation: R(φ, ψ) iff ` φ ↔ ψ.
Here, 1B = [p ∨ ¬p] and 0B = [p ∧ ¬p], for some p ∈ Σ. (It is important to note
that the negation symbol here is just the usual symbol in propositional logic –
crucially, this is not ¬B ). Also, we define [p] ∧B [q] = [p ∧ q], [p] ∨B [q] = [p ∨ q],
and ¬B [p] = [¬p].
v Make sure you see why these examples are indeed Boolean algebras! In particular,
it may be instructive to think of our first example when X contains two elements
and our last example where Σ contains one propositional constant. Do these
Boolean algebras look familiar?
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2. Boolean algebras: Homomorphisms.
• A homomorphism f : A → B, where A and B are Boolean algebras, is a map from
the elements of A to the elements of B which preserves the operations ∧, ∨, and ¬
(that is, f (¬A a) = ¬B f (a), for any a ∈ A, and so forth).
• There’s an interesting homomorphism from B30 to B6 . This homomorphism “flattens”
the cube into a diamond: g(x) = x/5 for x ∈ {5, 10, 15, 30}, and g(x) = x otherwise.
Can you check that g : B30 → B6 is in fact a homomorphism?
• Using the notion of a homomorphism, we can define an isomorphism between two
Boolean algebras A and B to be a bijective homomorphism f : A → B.
• We saw in class that some Boolean algebras are isomorphic to powersets with the
usual set-theoric operations. In class, too, the interesting question was posed: are all
Boolean algebras isomorphic to a powerset? As a practice problem, we will show that
this is not the case.
Proof. We know from set theory that, for any set X, the powerset of X has size 2X .
In particular, this shows that the smallest infinite powerset is uncountably infinite
(this is the case where X is countably infinite). So if we can show that there are
Boolean algebras with a countable number of elements, this will be enough to show
that these Boolean algebras can’t be isomorphic to a powerset. Consider one following
such Boolean algebra (as it turns out, the final example under 1. is also a countably
infinite Boolean algebra if Σ is countable, but we won’t prove this now): let B be the
set of finite and confinite sets of integers, where a cofinite set of integers is one that
omits only finitely many integers. The finite sets are clearly closed under finite union
and finite intersection, and likewise the cofinite sets are so closed. The complement
(with respect to the set of all integers) of a finite set is cofinite and vice versa. The
union of a finite set with a cofinite set is cofinite, while their intersection is finite.
Hence this set of sets of integers is closed under all the Boolean operations. One can
also show that these operations satisfy all the Boolean identities. So B is a countably
infinite Boolean algebra, as we wanted.
3. Boolean algebras: Practice problems.
Show that the following facts hold in an arbitrary Boolean algebra.
i 1 ≤ ¬a ∨ b iff a ≤ b.
ii a ∧ b = a iff a ∨ b = b.
iii a ∧ b ≤ c iff a ≤ ¬b ∨ c.
iv a 6= ¬a. (We assume that 0 6= 1.)
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