2
$\begingroup$

I have the following proof to verify without using truth tables but rather to use the laws or theorems of logical equivalence.

I am suppose to prove $(p\wedge q)\vee p\equiv p$, but I am stuck at$(p\wedge p)\vee (p\wedge q)\equiv p$ Distributive Law.

$\endgroup$

1 Answer 1

4
$\begingroup$

You've started out in a way that isn't really helpful. Instead of distributing the $p$, you want to instead factor out a $p$. (Remember this isn't quite like addition and multiplication; both of these operations distribute over each other.) So, write \begin{align*} (p \wedge ~q) \vee p &= (p \wedge ~q) \vee (p \wedge 1) \\ &= p \wedge (~q \vee 1) \\ &= p \wedge 1 \\ &= p \end{align*} and you're done.

$\endgroup$
1
  • $\begingroup$ Yes lol redid it and started with the commutative law first , and got it. Thanks for the answer @Mike $\endgroup$
    – Alec
    Commented Jun 16, 2016 at 20:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .