Proof
(related to Lemma: A Criterion for Valid Logical Arguments)
"If and only if" means that we have to show both implications "$\Rightarrow$" and "$\Leftarrow$" of the above lemma. Assume, our logical argument, which we will call "$\phi$" below, consist of the premises $p_1,\ldots,p_n$ and the conclusion $q$.
"$\Rightarrow$"
 Suppose that $\phi$ is a valid logical argument.
 Then, by definition a valid logical argument, the conclusion $q$ is true whenever all the premises $p_1,\ldots,p_n$ are simultaneously true, i.e. $$\text{ if }([[p_1]]_I=1\text{ and },\ldots,\text{ and }[[p_n]]_I=1),\text{ then } [[q]]_I=1$$ for all interpretations $I$.
 It follows from the truth table of the conjunction that $x=p_1\wedge,\ldots,\wedge p_n$ will be valued true: $[[x]]_I=1.$
 Furthermore, it follows from the truth table of the implication that if any other proposition $y$ is valued true $[[y]]_I=1,$ then the implication $x\Rightarrow y$ is also valued true: $[[x\Rightarrow y]]_I=1.$
 In our case, $x\Rightarrow q$ and $q$ is true by assumtion. Therefore, the proposition $x\Rightarrow q$ is always valued true.
 Thus, $(p_1\wedge\ldots\wedge p_n)\Rightarrow q$ is a tautology.
"$\Leftarrow$"
 Conversely, suppose that $(p_1\wedge\ldots\wedge p_n)\Rightarrow q$ is a tautology, but the $\phi$ is fallacy.
 This would mean that $q$ is false, even though the premises $p_1,\ldots,p_n$ are all true.
 With the above notation this means that $q$ is false, even though $x=(p_1\wedge\ldots\wedge p_n)$ is true.
 From the truth table of implication, it follows that $x\Rightarrow q$ is false.
 But in this case, $(p_1\wedge\ldots\wedge p_n)\Rightarrow q$ cannot be a tautology, which is a contradiction.
 Therefore, $\phi$ is not a fallacy, but it must be a valid logical argument.
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References
Bibliography
 Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016
 Kohar, Richard: "Basic Discrete Mathematics, Logic, Set Theory & Probability", World Scientific, 2016