Question

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Answer 1

What exactly are you trying to prove? Equivalence? Implication? There's a missing symbol somewhere, and I'm guessing it's here:
\[p\land(q\lor r)\color{red}{\boxed{\color{white}0}}(p\land q)\lor(p\land r)\]

Answer 2

In case it's not clear, I'm using \(\land\) for "and" and \(\lor\) for "or" (which is standard notation, hope it's not confusing)

Answer 3

sorry, my bad, it didnt copy.

it is the 'is the same as' sign.. (equal sign with 3 dashes)

Answer 4

also the ! should be an arrow instead (to show p as antecedent and '(q∨q)' as consequent

Answer 5

Okay, thanks. Truth tables are a good place to start. You have three variables, so there are \(2^3=8\) cases to consider.
\[\begin{array}{c|c|c|c|c}
p&q&r&q\lor r&\color{red}{p\land (q\lor r)}&p\land q&p\land r&\color{red}{(p\land q)\lor(p\land r)}\\
\hline
1&1&1\\
1&1&0\\
1&0&1\\
1&0&0\\
0&1&1\\
0&1&0\\
0&0&1\\
0&0&0
\end{array}\]where \(1\) means "true" and \(0\) means "false". For the first statement to be true, both columns marked red must match truth values.

All you need to fill in the table is knowledge of the \(\land\) and \(\lor\) operators:
\[\begin{array}{c|c|c}
\land&1&0\\
\hline
1&1&0\\
\hline
0&0&1
\end{array}\quad\quad\quad\quad \begin{array}{c|c|c}
\lor&1&0\\
\hline
1&1&1\\
\hline
0&1&0
\end{array}\]For example, the fourth and fifth columns would get filled like so:
\[\begin{array}{c|c|c|c|c}
p&q&r&q\lor r&\color{red}{p\land (q\lor r)}&p\land q&p\land r&\color{red}{(p\land q)\lor(p\land r)}\\
\hline
1&1&1&1&\color{red}1\\
1&1&0&1&\color{red}1\\
1&0&1&1&\color{red}1\\
1&0&0&0&\color{red}0\\
0&1&1&1&\color{red}0\\
0&1&0&1&\color{red}0\\
0&0&1&1&\color{red}0\\
0&0&0&0&\color{red}0
\end{array}\]

Answer 6

Just saw your second comment. The above procedure works for establishing (or disproving) equivalence.

To prove an implication, recall that an implication is true if either both the antecedent and consequent are true, or the antecedent is false:
\[\begin{array}{c|c|c}
\implies&1&0\\
\hline
1&1&0\\
\hline
0&1&1
\end{array}\]