Show that $(p \rightarrow q) \wedge (p \rightarrow r)$ and $p \rightarrow (q \wedge r)$ are logically equivalent.

I tried to do this making a truth table but I think my teacher wants me to solve it using the different laws of Logical Equivalences.

Can anyone help me?

  • 4
    $\begingroup$ Which equivalences do you have available? Particularly involving $\to$. $\endgroup$ – hmakholm left over Monica Mar 15 '15 at 16:44
  • $\begingroup$ I have that $p \rightarrow q \equiv \neg p \vee q$ $\endgroup$ – Aziz Soldier Mar 15 '15 at 16:47
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    $\begingroup$ x @Aziz: Use that on both sides, then the distributive law. $\endgroup$ – hmakholm left over Monica Mar 15 '15 at 16:51
  • $\begingroup$ If I do it I'll get $\neg p \vee q \wedge \neg p \vee r$ then $\neg p \vee (q \wedge r)$ and $p \rightarrow (q \wedge r)$ and it is correct. Thank you @ henning Makholm $\endgroup$ – Aziz Soldier Mar 15 '15 at 16:58

Here is an approach

$$(p \to q) \wedge (p \to r) \equiv (\neg p \vee q) \wedge (\neg p \vee r) \equiv \neg p \vee (q \wedge r) \equiv p \to (q \wedge r)$$


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