Uncategorized – Page 3 – MAT2440 Discrete Structures & Algorithms I

Prove that the following compound propositions are equivalent:

$\neg (p \wedge q) \equiv \neg p \vee \neg q$
$\neg (p \vee q) \equiv \neg p \wedge \neg q$
$p \to q \equiv \neg p \vee q$
$p \to q \equiv \neg q \to \neg p$
$p \vee q \equiv \neg p \to q$
$p \wedge q \equiv \neg (p \to \neg q)$
$\neg(p \to q) \equiv p \wedge \neg q$
$(p \to q) \wedge (p \to r) \equiv p \to (q \wedge r)$
$(p \to r) \wedge (q \to r) \equiv (p \vee q) \to r$
$(p \to q) \wedge (p \to r) \equiv p \to (q \vee r)$
$(p \to r) \wedge (q \to r) \equiv (p \wedge) \to r$
$(p \vee q) \vee r \equiv r \equiv p \vee (q \vee r)$ and $(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)$
$p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)$ and $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$
$p \vee (p \wedge q) \equiv p$ and $p \wedge (p \vee q) \equiv p$
$p \wedge q \equiv q \wedge p$ and $p \vee q \equiv q \vee p$
$T \wedge p \equiv p$ and $F \vee p \equiv p$
$T \vee p \equiv T$ and $T \wedge p \equiv p$

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Category: Uncategorized (Page 3 of 3)

1.3 Propositional equivalences – in-class exercises

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