# propositional logic

• August 30th 2011, 08:48 AM
llapkan
propositional logic
hey friends,
show that (p→q)∧(p→r) and p→(q∧r) is logically equivalent

• August 30th 2011, 09:06 AM
Plato
Re: propositional logic
Quote:

Originally Posted by llapkan
show that (p→q)∧(p→r) and p→(q∧r) is logically equivalent
$\begin{gathered} (p \to q) \wedge (p \to r) \equiv \hfill \\ (\neg p \vee q) \wedge (\neg p \vee r) \equiv \hfill \\ \neg p \vee (q \wedge r) \equiv p \to (q \wedge r) \hfill \\ \end{gathered}$
Another way to do this is to construct a "truth table". p, q, and r can each be "T" or "F" so there are $2^3= 8$ cases.
If p= q= r= T, then $p\to q$ is true and $p\to r$ is true so $(p\to q)\wedge(\p\to r)$ is true. Also then [tex]q\wedge r[/itex] is true so $p\to (q\wedge r)$ is true. Since they are both true $(p\to q) \wedge (p\to r)= p\to (q \wedge r)$ in this case.
If p= F, which includes 4 of the 8 cases, each of $p\to q$, $p\to r$, and $p\to(q\wedge r)$ is true because " $F\to x$" is true no matter what x is.