Proofs

**captainjapan** · October 17th 2008, 01:14 PM

Prove by direct proof, using only equivalence and inference rules. Remember to name each rule used.
1. (p ^ q) → r

2. p ^ ¬ r

Therefore ¬q.

**Plato** · October 17th 2008, 02:32 PM

$\begin{gathered} \left( {p \wedge q} \right) \to r \hfill \\ p \wedge \neg r \hfill \\ \------- \hfill \\ p \hfill \\ \neg r \hfill \\ \neg \left( {p \wedge q} \right) \hfill \\ \neg p \vee \neg q \hfill \\ \neg q \hfill \\ \end{gathered}$
You fill in the reasons.

**Soroban** · October 17th 2008, 02:56 PM

Hello, captainjapan!

$(p \to q) \;\Leftrightarrow\;\sim p \vee q$ . Alternate definition of implication (ADI)

Prove by direct proof, using only equivalence and inference rules.
Remember to name each rule used.

$\begin{array}{c}(p \wedge q) \to r \\ p \:\wedge \sim\! r \\ \hline \therefore\;\sim\! q \end{array}$

. . $\begin{array}{ccc} (p \wedge q ) \to r & & \text{Given} \\ \sim(p \wedge q) \vee r & & \text{ADI} \\ (\sim\!p \:\vee \sim\!q) \vee r & & \text{DeMorgan} \\ (\sim\! p \vee r) \:\vee \sim\! q & & \text{comm/assoc} \\ \sim(\sim\!p \vee r) \to \:\sim\!q & & \text{ADI} \\ (p \:\wedge \sim\!r) \to \:\sim\! q & & \text{DeMorgan} \\ p \:\wedge \sim r & & \text{Given} \\ \therefore\;\;\sim\!q & & \text{Modus Ponens} \end{array}$

**Plato** · October 17th 2008, 03:18 PM

Originally Posted by Soroban

$(p \to q) \;\Leftrightarrow\;\sim p \vee q$ . Alternate definition of implication (ADI)

In almost all standard logic textbooks the property
$(p \to q) \;\Leftrightarrow\;\sim p \vee q$ is known as Material Implication (Impl.).

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