1.Determine whether ¬(p∨(¬p∧q)) and ¬p∧¬q are equivalent without using truth table. 2.Determine whether the compound proposition ~(p∨q)∨(~p∧q)∨p to tautology. 3.Determine whether (p → q)∧(p → r)≡p → (q ∧r) using a truth table

Let T denotes True and F denotes False.

1.Let us determine whether $¬(p∨(¬p∧q))$ and $¬p∧¬q$ are equivalent:

$¬(p∨(¬p∧q))=¬p\land\neg(¬p∧q))=¬p\land(p\lor\neg q))=$

$=¬p\land p\lor ¬p\land\neg q=F\lor ¬p\land\neg q= ¬p\land\neg q$ .

Therefore, the formulas are equivalent.

2.Let us determine whether the compound proposition $\sim(p∨q)∨(\sim p∧q)∨p$ is tautology.

$\sim(p∨q)∨(\sim p∧q)∨p=\sim(p∨q)∨(\sim p\lor p)∧(q∨p)=\sim(p∨q)∨T∧(q∨p)=$

$=\sim(p∨q)∨(q∨p)=\sim(p∨q)∨(p\lor q)=T.$

Therefore, the formula $\sim(p∨q)∨(\sim p∧q)∨p$ is tautology.

3.Let us determine whether (p → q)∧(p → r) ≡ p → (q ∧r) using a truth table:

$\begin{array}{||c|c|c||c|c|c|c|c||} \hline\hline p & q & r & p\to q & p\to r & (p\to q)\land (p\to r) & q\land r & p\to(q\land r)\\ \hline\hline F & F & F & T & T & T & F & T\\ \hline F & F & T & T & T & T & F & T \\ \hline F & T & F & T & T & T & F & T \\ \hline F & T & T & T & T & T & T & T \\ \hline T & F & F & F & F & F & F & F\\ \hline T & F & F & F & T & F & F & F \\ \hline T & T & F & T & F & F & F & F\\ \hline T & T & T & T & T & T & T & T\\ \hline\hline \end{array}$

Since the formulas $(p\to q)\land (p\to r)$ and $p\to(q\land r)$ always have the same truth values, they are logically equivalent.