🗒️ Ben's Notes

Propositional Logic

What are Propositions? #

Propositions are anything that can be true or false. This could include:

  • Statements like “Birds can fly”.
  • Well defined equations with no free variables like $1 + 1 = 3$.

Propositions are not:

  • Variables like $x$ or $5$.
  • Equations with free variables like $P(x) = y$.
  • Statements that aren’t clearly true or false, like “I like trains.”

Connectives #

Simple propositions can be joined together to make complex statements. There are three basic ways to connect propositions together:

  • Conjunction is the and operation: for $P \land Q$to be true, $P$and $Q$must both be true.
  • Disjunction is the or operation: for $P \lor Q$to be true, either $P$or $Q$must be true.
  • Negation is the not operation: if $P$is true, then $\lnot P$is false.
    • The law of the excluded middle states that $P$and $\lnot P$ cannot both be true.

One example where we can see these components in action is in De Morgan’s Laws, which state how negation can be distributed across conjunction or disjunction:

$ \lnot(P \lor Q) \iff (\lnot P \land \lnot Q) $

“If neither P nor Q are true, then P and Q must both be false.”

$ \lnot(\forall x)(P(x)) \iff (\exists x)(\lnot P(x)) $

“If P(x) isn’t true for every x, then there exists an x where P(x) is false.”


Another example of distribution is this congruence, which works for any combination of and’s and or’s.

$ (P \lor Q) \land R \equiv (P \land R) \lor (Q \land R) $


Implication #

One proposition can imply another, which looks like this:

$ P \implies Q $

Roughly, implication in plain English can be stated in the form if P, then Q. However, there are a lot of nuances to what this really means!

Properties of Implication #

  • Reversible: Q is true if P is true. However, be careful- this doesn’t necessary mean that Q implies P!
  • P is sufficient for Q: Proving P allows us to say that Q is also true.
  • Q is necessary for P: For P to be true, it is necessary that Q is true. (If Q is false, then P is also false.)
  • Contrapositive Equivalence: If P implies Q, then $\lnot Q \implies \lnot P$.
    • Note that this is different from the converse, which is $Q \implies P$. This statement is not logically equivalent!

Truth Table #

PQP $\implies$QP $\iff$Q
TTTT
TFFF
FTTF
FFTT

Note that the truth table for $P \implies Q$ is equivalent to the one for $\lnot P \lor Q$! That means this formula is logically the same as $P \implies Q$.

(If two propositions have the same truth table, then they are logically equivalent. However, it’s still possible for a proposition to imply another even if their truth tables are different!)

Quantifiers #

Sometimes, we need to define a specific type of variable to work with in a propositional clause. For instance, take the proposition, “There exists a natural number that is equal to the square of itself.” We could write this as:

$ (\exists x \in \mathbb{N})(x=x^2) $

You could think about the parentheses almost like defining a scope of variables, like what might happen in programming! Here, the first clause is defining an arbitrary variable $x$to be any natural number.

Exercises #

Is the expression $\forall x \exists y (Q(x,y) \implies P(x))$equivalent to the expression $\forall x ((\exists y \\ Q(x,y)) \implies P(x))$?
(Source: Discussion 0 2a)

No, they are not equivalent. We can see this more clearly by converting the implication $Q \implies P$ to $\lnot Q \lor P$ as was demonstrated in the Truth Table section above.

On the left side, this conversion is straightforward, yielding $\forall x \exists y (\lnot Q(x,y) \lor P(x))$.

On the right side, we’ll need to invoke De Morgan’s Laws to convert the ’exists’ into a ‘for all’ since it was negated. This yields $\forall x (\forall y\lnot(Q(x,y)) \lor P(x))$which is not the same thing!

An integer $a$is said to divide another integer $b$ if $a$is a multiple of $b$. Write this idea out using propositional logic (a divides b can be written as $a \mid b$).

Note: This idea is going to be important for a lot of future sections!

$a \mid b \iff (\exists q \in \mathbb{Z})(a = qb)$

In plain English: “There exists an integer $q$such that when we multiply $q$with $b$, we get $a$.”

Resources #

Note 1: https://www.eecs70.org/assets/pdf/notes/n1.pdf
Discussion 0: https://www.eecs70.org/assets/pdf/dis00a.pdf