Logical equivalences

**captainjapan** · September 29th 2008, 02:12 PM

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)

**Soroban** · September 29th 2008, 03:09 PM

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).

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

**Plato** · September 29th 2008, 03:16 PM

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)

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