Validate an argument with if and only if How would I validate this argument?
$p \iff  q$
$r \vee q$
$\neg r$
$\overline{\therefore \neg p\quad}$    
Is this Valid or Invalid?
I would say this argument is invalid, because r or q doesn't mean if and only if
 A: You are indeed correct that the argument is not valid.
Premises:
$p \leftrightarrow q \equiv \color{blue }{\bf (p \rightarrow q) \land (q\rightarrow p)}\tag{1}\\$
$r \lor q \equiv \color{red}{\bf \lnot r \rightarrow q}\tag{2}\\$
$\color{red}{\bf \lnot r}\tag{3}$
From the second premise: we can use either the left-hand-side of $(2)$, or the right-hand-side. If you're more comfortable with modus ponens, we can use the right hand side: $\color{red}{\bf \lnot r \rightarrow q}$. Then with the third premise: $\color{red}{\bf \lnot r},\;$ we can conclude $\color{red}{\bf  q}$, by modus ponens.
But note: from the left-hand-side of premise $(2)$: $\;\bf{r \lor q}\,$ along with the third premise $\,\bf{\lnot r},\,$ we can derive/deduce/infer $\,\bf q\;$ by invoking the valid rule-of-inference sometimes called the disjunctive syllogism.
So we have $\color{blue}{\bf q},\,$ and from simplification of the first premise, we have $\color{blue}{\bf q\rightarrow p}$.
From $\color{blue}{\bf q \rightarrow p}, \;\text{and}\;\color{blue}{\bf q},\,$ we therefore have the valid inference $\,\color{blue}{\bf p},\,$ by modus ponens again.
So you are correct that the conclusion $\,\lnot p\,$ is invalid
A: $r\vee q$ tells you that at least one of $r,q$ is true,$\neg r$
tells that $r$ is false, hence $q$ is true.
$p\iff q$ tells you that either both of $p,q$ are true or both are
them are false, but $q$ is true and hence so is $p$.
A: Since $(r \vee q) \wedge \neg r$, it must be the case that $q$ is true.
Then, since $(p \iff q) \wedge q$, it must be the case that $p$ is true.
Thus, the argument that the premises imply $\neg p$ is invalid.
A: If an argument qualifies as valid, then if all of the premises qualify as true, so will its conclusion.  If an argument does not qualify as true, then all of its premises can hold true, while its conclusion may qualify as false.
So, suppose all premises here true.
Thus, ¬r qualifies as true.  So, r qualifies as false.  Since r qualifies as false, and (r∨q) qualifies as true, this implies q as true.  Since (p⟺q) qualifies as true also by assumption, then p qualifies as true.  But, then ¬p qualifies as false.  Consequently, the argument is not valid.
