Can we prove an ‘if’ statement by proving that its converse is false? That is, is $$\lnot (B \implies A); \text{ therefore, } (A \implies B)$$ a sound argument? Speaking to several colleagues about this has not convinced me of this. The sentence $$\lnot (B \rightarrow A) \to (A \rightarrow B)$$ is certainly valid, as can be seen from its truth table. However, its antecedent implies $\lnot A$ which feels like it should violate soundness.
Logic – Proving an Implication by Disproving Its Converse
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I see that valid formula (proposition or statement) is the one that is valid under every interpretation. But this is a tautology. Is there any difference between tautology and valid formula?
Some people use "tautology" in logic in a wide sense, to mean any logically true wff. But others use "tautology" more narrowly to mean true in virtue of truth-functional structure (so, "valid by the truth-table test"). So, for example, $\forall xFx \to Fa$ would count as a tautology in the first, wide, sense but not in the second, narrow, sense. (How to use 'tautology', then, is a matter of terminological preference: I much prefer the second narrower usage as it gives us a label for a special class of logical truths which we need a label for.)
They also say that $A, B \vDash \Phi$ is valid if it preserves the truth
No they don't. Or at least, they shouldn't. It is the inference that preserve truth. Thus if $A, B \therefore \Phi$ necessarily preserves truth, then it is valid. But $A, B \vDash \Phi$ is not an inference: it is a meta-linguistic statement about the inference, and says the inference is valid.
They also say that in addition to being valid, i.e. preserving the truth, your argument is sound when premises are also valid.
Fine: an argument is sound if it has true premisses and a valid inference move.
But there is another definition of soundness, which tells that theory is sound if formula provability implies that formula is semantically valid.
As you almost say, this is a definition of what it is for a formal deductive system to be sound.
Arguments aren't formal deductive systems, and soundness for arguments is not soundness for formal deductive systems. The notions can't be identical. But there is some similarity between the two notions -- both have something to do with being trustworthy (an argument won't lead you astray about the truth if it starts with truths and proceeds by truth-preserving moves, i.e. is sound; likewise a formal system encoding inferences won't lead you astray about what's a valid inference if it is sound)
The discussion is about why the statement $\bot \to \bot$ is considered "true" rather than "false". That is, why the truth table of the conditional connective is defined as it is.
An argument is considered valid if, it guarantees the conclusion is true when all the premises are true.
So if $\to$ is defined as it is, then the truth of both premises, $\{P\to Q, P\}$ are required to guarantee the truth of the conclusion $Q$. (Knowing only that one is true is not enough to ensure the conclusion is true.)
$$\begin{array}{cc|cc|cc}P & Q & P\to Q & P & Q \\ \hline \bot & \bot & \top & \bot & \bot \\ \bot & \top & \top & \bot & \top \\ \top & \bot & \bot & \top & \bot \\ \top & \top & \top & \top & \top & \star \end{array}$$
However if we defined $\dot\to$ so that $\bot \dot\to \bot = \bot$ we only need the one premise $P\dot\to Q$ to guarantee the conclusion. The conclusion is always true when that premise is, whatever $P$ may be.
$$\begin{array}{cc|c|cc}P & Q & P\dot\to Q & Q \\ \hline \bot & \bot & \bot & \bot \\ \bot & \top & \top & \top & \star \\ \top & \bot & \bot & \bot \\ \top & \top & \top & \top & \star \end{array}$$
Best Answer
Proving an implication by disproving its converse is legitimate only if said implication is not quantified.
Logically, the non-quantified statement $$\text{Amy being a vegan implies that she eats beef}$$ and its converse $$\text{Amy eating beef implies that she is a vegan}$$ cannot both be false: if one is false, then the other must be vacuously true.
However, the quantified statement $$\text{Every vegan eats beef}$$ and its converse $$\text{Every beef-eater is a vegan}$$ are both false.
Explanation
The open formula $A(x)\rightarrow B(x)$ often implicitly means the sentence $\forall x\,\Big(A(x)\rightarrow B(x)\Big),$ which is different from the sentence $A \rightarrow B.$
To be clear: your suggested proof/argument's premise $\lnot (B \rightarrow A)$ is satisfiable, regardless of the fact that it is consistent with your target implication $(A\to B)$'s hypothesis $A$ being false. On the other hand, if you reframe your proof as having premise $\big(\lnot (B \rightarrow A)\;\land\;A\big)$ and conclusion $B,$ then this premise is still consistent with your observation that $\lnot (B \rightarrow A)\to\lnot A$ is a tautology. In both cases, your suggested argument isn't categorically unsound. It is unsound precisely when $A$ is true and $B$ false, in which case your target implication (is false and) cannnot be proven.