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# Propositional Logic

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."

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.

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

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!**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!**

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!)

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.Q1

Answer 1

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!Q2

Answer 2

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 $a$

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

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