1. ## Logical equivalences

Use the logical equivalences to show that:

(a) (¬p → (q → r)) ≡ (q → (p V r))

(b) ¬(p → ¬q) & ¬(p V q) is a contradiction (i.e. always false).

(c) (p V q) & (¬p V r) → (q V r) is a tautology (i.e. always true)

2. Hello, captainjapan!

$(a \to b) \;\equiv\: \sim a \vee b$ . I call it ADI (alternate definition of implication).

$(a)\;\;[\sim p \to (q \to r)] \: \equiv\: [q \to (p \vee r)]$
On the left side we have:

. . $\begin{array}{ccc}\sim p \to (q \to r) & & \text{Given} \\ \\ [-4mm]

p \vee (q \to r) & & \text{ADI} \\ \\ [-4mm]

p \vee (\sim q \vee r) & & \text{ADI} \\ \\ [-4mm]

\sim q \vee (p \vee r) & & \text{comm., assoc.} \\ \\[-4mm]

q \to (p \vee r) & & \text{ADI} \end{array}$

$(b)\;\;\sim (p \to \; \sim q)\: \wedge \sim(p \vee q)$ is a contradiction (i.e. always false).

. . $\begin{array}{ccc}
\sim(p \to \:\sim q)\; \wedge \sim(p \vee q) & & \text{Given} \\ \\[-4mm]
\sim(\sim p \:\vee \sim q) \;\wedge \sim(p \vee q) & & \text{ADI} \\ \\[-4mm]
(p \wedge q) \wedge (\sim p \:\wedge \sim q) & & \text{DeMorgan} \\ \\[-4mm]
(p \:\wedge \sim p) \wedge (q \:\wedge \sim q) & & \text{comm, assoc.} \\
f \wedge f \\ \\[-4mm]
f
\end{array}$

3. Originally Posted by Soroban
$(a \to b) \;\equiv\: \sim a \vee b$ . I call it ADI (alternate definition of implication).
FYI: In formal logic it is called Material Implication (Impl)