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If I have a statement that says,

$I\implies J\implies K$.

Then

Can I say that,

since, $J \implies K$ and $J$ is true, $K$ is true as well.

Or Do I have to prove that $I$ is true as well to say so?

2 Answers 2

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Sometimes we briskly express the three metalogical claims that (1) implies (2), and (2) implies (3), and (3) implies (1) in the compressed form $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$ So in this case, to be sure, if e.g. you already have an independent proof that (2) is true, you could extract the claim that $(2) \Rightarrow (3)$ and conclude that (3).

However we never make that contraction inside the propositional calculus: i.e. we never contract $p \to q \land q \to r$ to $p \to q \to r.$ Depending on your official rules for dropping brackets, the latter is either illegitimate or is to be read as $p \to (q \to r)$ which means something quite different. And to conclude $r$ given the latter, you'd need to invoke both the truth of $p$ and the truth of $q$.

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    this one is more convincing. thanks2012-11-29
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$J\Rightarrow K$ means that when $J$ is true, $K$ must be true as well, so, yes, this would be a reasonable deduction given $J$.

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    I took this to mean $I \Rightarrow J$ and $J \Rightarrow K$, but you raise a good point. There was some ambiguity in the question and$I$answered quickly without considering this alternative viewpoint.2012-11-28