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."

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

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

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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

P	Q	P $\implies$Q	P $\iff$Q
T	T	T	T
T	F	F	F
F	T	T	F
F	F	T	T

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

Q1

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)

Answer 1

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!

Q2

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!

Answer 2

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

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