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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