Determine if Modus Ponens argument is valid I am trying to determine whether the following argument is valid:
P ⇒ ((∼Q) ∧ R)
Q ⇒ (P ∨ R)
therefore P ⇒ R
I have constructed truth tables for each statement. However, I am confused on how to proceed as I have only seen Modus Ponens in its most basic definition.
 A: If you know that $$P \implies (\neg Q \wedge R),$$
then in particular you know
$$P \implies R.$$
A: Instead of making a separate truth-table for each statement, you need to create a combined truth-table for all. That truth-table should look like this:
\begin{array}{c|c|c|c|c|c}
P&Q&R&P \to (\neg Q \land R)&Q \to (P \lor R)&P \to R\\
\hline
T&T&T&F&T&T&\\
T&T&F&F&T&F&\\
T&F&T&T&T&T&\\
T&F&F&F&T&F&\\
F&T&T&T&T&T&\\
F&T&F&T&F&T&\\
F&F&T&T&T&T&\\
F&F&F&T&T&T&\\
\end{array}
OK, but does this mean the argument is valid or invalid?  Well, an argument is valid if and only if the truth of the premises implies the truth of the conclusion. In other words: an argument is valid if and only if the conclusion is true whenever the premises are true. Or differently yet: an argument is valid if and only if it is impossible for the conclusion to be false whenever the premises are true. Or more simply:  an argument is valid if and only if it is impossible to have true premises and a false conclusion.
If you take that last way of looking at validity, then that tells you what to look for in the truth-table: a row where all premises are true but the conclusion is false. If there is such a row, then apparently it is possible to have true premises and a false conclusion, and hence the argument is not valid. If, on the other hand, there is no such row, then apparently it is not possible to have true premises and a false conclusion, and hence the argument is valid.
Well, we have 8 rows in the truth-table, but in none of them do we have true premises and a false conclusion. Hence, the argument is valid.
A: We want to prove $P \to (\neg Q \land R), Q \to (P \lor R) \therefore P \to R\,$ is a valid argument. If we can derive this conclusion 
$$P \to R$$ 
from the premises $$P \to (\neg Q \land R), Q \to (P \lor R)$$ we can assert that it is a valid argument.
As we want to prove a sentence whose main logical connective is $\mathbf{\to}$, we would use the rule $\mathbf{\to I}$ (Conditional Introduction). We make an additional assumption, $P$; and from that additional assumption, we prove $R$.
$
\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}
\def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\}
\def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\}
\def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\}
\def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\}
\def\R#1{\qquad\mathbf{R} \: #1 \\}
\def\ci#1{\qquad\mathbf{\land I} \: #1 \\}
\def\ce#1{\qquad\mathbf{\land E} \: #1 \\}
\def\ii#1{\qquad\mathbf{\to I} \: #1 \\}
\def\ie#1{\qquad\mathbf{\to E} \: #1 \\}
\def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\}
\def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\}
\def\qi#1{\qquad\mathbf{=I}\\}
\def\qe#1{\qquad\mathbf{=E} \: #1 \\}
\def\ne#1{\qquad\mathbf{\neg E} \: #1 \\}
\def\ni#1{\qquad\mathbf{\neg I} \: #1 \\}
\def\IP#1{\qquad\mathbf{IP} \: #1 \\}
\def\x#1{\qquad\mathbf{X} \: #1 \\}
\def\DNE#1{\qquad\mathbf{DNE} \: #1 \\}
$
$
\fitch{1.\, P \to (\neg Q \land R) \\ 2.\,Q \to R}{
 \fitch{3.\, P}{
 4.\, \neg Q \land R \ie{1,3}
 5.\, R \ce{4}
}\\
6.\, P \to R \ii{3-5}
}
$
P.S.: $\mathbf{\to E}$ rule is commonly known as Modus Ponens.
A: $\color{red} {\underline{\overline{\ First \space Method : Using \space a \space truth-table }}}$
You can find out the answer by first writing  the corresponding conditional of the reasoning, that is : 

(Premise $1$ & Premise $2$ ) $\rightarrow$ Conclusion

and then building its truth table ( using for example Mickael Rippel's Truth-table Generator  : https://mrieppel.net/prog/truthtable.html)

The reasoning is valid iff this corresponding conditional is a
  tautology ( = a formula that is true in all possible cases).

Note : what has to be a tautology ( in order the reasoning to be valid ) is the whole big conditional, not the premises by themselves nor the conclusion by itself. 
Below , the truth-table I obtained from Mickael Rippel's site : 

$\color{red} {\underline{\overline{\ Second \space Method : Trying \space to \space find \space a  \space counter-example. }}}$


*

*Suppose ( for refutation) this reasoning is  not valid. 


$(1)$ It means there is a possible case in which  the premises are true and the conclusion is false. ( For, a reasoning is valid if and only if  there is no possible case in which the conclusion is false while all the premises are true: this is the standard definition off " validity"). 
$(2)$ If the conclusion , namely $(P \rightarrow R)$ , is false ( hypothetically) then $P$ is true and $R$ is false ( this is the only way for  this conditional to be false). 
$(3)$ Now your premises are true ( by hypothesis) , and in particular $ P \rightarrow ( \neg Q \land R)$ is true. 
This conditional has a true antecedent ( since P is true, as we said). 
The only way for a conditional with a true antecedent to be true is to have a true consequent. So $( \neg Q \land R)$ must be true. 

$(4)$ Furthermore, the only way for a conjunction to be true is to
  have both its conjuncts true. So , in particular, $R$ is true. 
$(5)$ But we previously admitted ( step #2 ) that $R$ is false. 
So, our hypothetical case in which the premises are true and the
  conclusion false is a contradctory case, an impossible case.

$(6)$ We can conclude that : there is no possible case in which the conclusion is false while the premises are true; which amounts to saying that the reasoning is valid. 
Note : more about this method in Mendelson, Outline Of Boolean Algebra. 
A: You don't need to make three different truth tables, just one:

As you can see, it is a tautology. Using a form of natural deduction (in DC Proof 2.0), we have (where '|' = OR):

