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Math Help - Tautology

  1. #1
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    Tautology

    (p V q) & (p V r) → (q V r) is a tautology (i.e. always true)
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  2. #2
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    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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