# Lecture 005

## Connectives

Biconditional Statements($\iff$, $\Leftrightarrow$, $\equiv$, $\leftrightarrow$)

• "P is and only if Q"

• "P is logically equivalent to Q"

• $(P \implies Q)\land(Q\implies P)$

P Q P->Q Q->P P<->Q
T T T T T
T F F T F
F T T F F
F F T T T

DeMorgan's Laws:

• tautology: a proposition that is always true for all possible truth assignments.

• DeMorgan's Law for Connectives: "For all propositions P and Q, the following logical equivalences holds true"

• $\lnot(P\land Q)\equiv \lnot P\land\lnot Q$
• $\lnot(P\lor Q)\equiv \lnot P\lor\lnot Q$
• DeMorgan's Laws for Sets: "For any sets A and B with a universal set U, the following hold true."

• $\overline{A \cap B} = \overline{A} \cup \overline{B}$
• $\overline{A \cup B} = \overline{A} \cap \overline{B}$
• Proof: because $\lnot(x\in A \land x\in B)\iff \lnot(x\in A)\lor \lnot(x\in B)$ due to connectives
P Q $P \land Q$ $\lnot(P \land Q)$ $\lnot P \lor \lnot Q$ $\lnot(P\land Q) \iff \lnot P \lor \lnot Q$
T T T F F T
T F F T T T
F T F T T T
F F F T T T

Distributive Law for Connectives:

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

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

you can think this as distributive

Law of Double Negation: $\lnot \lnot P \equiv P$

Disjunctive Form of Implication: $P\implies Q\equiv \lnot P \lor Q$

Disjunctive Form of Implication: $P\iff Q \equiv (P\implies Q)\land (Q\implies P)$

Contraposition(contrapositive): $P\implies Q \equiv \lnot Q \implies P$

• converse: $Q\implies P$ is converse of $P\implies Q$

• inverse: $\lnot P \implies \lnot Q$ is inverse of $P\implies Q$

• converse and inverse are logically equivalent to each other (since they are contraposition to each other)

Law of Excluded Middle: $P(x) \neq \lnot P(x)$

## Maximally Negating Propositions

\lnot(\forall x \in \mathbb{R})(\exists y \in \mathbb{R})(xy=1) \equiv (\exists x \in \mathbb{R})(\forall y \in \mathbb{R})(xy\neq 1)
• both our qualifiers changed

• right (inner) most variable proposition negated

Maximally Negated Form:

• $\lnot(\forall x \in S)P(x)\equiv(\exists x \in S)\lnot P(x)$

• $\lnot(\exists x \in S)P(x)\equiv(\forall x \in S)\lnot P(x)$

\lnot \alpha \equiv \lnot(\forall x \in S)(\exists y \in S)(\forall z \in S) (P (x) \land Q(y)) \implies R(x, y, z)\lnot \alpha \equiv (\exists x \in S)(\forall y \in S)(\exists z \in S) (P (x) \land Q(y)) \land \lnot R(x, y, z)

Notice that S, P(x), Q(y) does not change

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