Tautology

**captainjapan** · October 1st 2008, 05:07 AM

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

**Soroban** · October 1st 2008, 06:17 AM

Hello, captainjapan!

Once again, I apply my own names to defintions and axioms.

. . $\begin{array}{ccc}p \to q \;=\;\sim p \vee q & & \text{ADI} \\ \\[-4mm]<br /> p \:\vee (\sim p) \;=\;t & &\text{axiom 1} \\ \\[-4mm]<br /> p \wedge t \;=\;p & & \text{axiom 2} \\ \\[-4mm]<br /> p \vee t \;=\;t & & \text{axiom 3} \end{array}$

Prove that: . $[(p \vee q) \wedge (\sim p \vee r)] \to (q \vee r)$ .is a tautology.

. . . . . $[(p\vee q) \wedge )\sim p \vee r)] \to(q \vee r)$ . . . . . . . . Given

. . . . $\sim\bigg[(p \vee q) \wedge (\sim p \vee r)\bigg] \vee (q \vee r)$ . . . . . . . . ADI

. . . . $\sim(p \vee q) \:\vee \sim(\sim p \vee r) \vee q \vee r$ . . . . . . . DeMorgan

. . . . $(\sim p \:\wedge \sim q) \vee (p \:\wedge \sim r) \vee q \vee r$ . . . . . . . DeMorgan

. . . $\bigg[(\sim p \:\wedge \sim q) \vee q\bigg] \vee \bigg[(p \:\wedge \sim r) \vee r\bigg]$ . . . . . Comm, Assoc.

$\bigg[(\sim p \vee q) \wedge (\sim q \vee q)\bigg] \vee \bigg[(p \vee r) \wedge (\sim r \vee r)\bigg]$ . . Distributive

. . . . . $\bigg[(\sim p \vee q)\wedge t\bigg] \vee \bigg[(p \vee r) \wedge t\bigg]$ . . . . . . . . axiom 1

. . . . . . . . . $[\sim p \vee q] \vee [p \vee r]$ . . . . . . . . . . . axiom 2

. . . . . . . . . $(q \vee r) \vee (\sim p \vee p)$ . . . . . . . . Comm, Assoc.

. . . . . . . . . . . $(q \vee r) \vee t$ . . . . . . . . . . . . . axiom 1

. . . . . . . . . . . . . . $t$ . . . . . . . . . . . . . . . .axiom 3

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