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If I have a statement that says,

$I\implies J\implies K$.


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?

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up vote 0 down vote accepted

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. thanks – user45099 Nov 29 '12 at 8:51

$J\Rightarrow K$ 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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But does not I=>J=>K mean I=>(J=>K). But we do not know if I is true!!! So we do not know if J=>K is true, right? – user45099 Nov 28 '12 at 18:55
If I were not true, $I \implies (J \implies K)$ would be true vacuously (in classical logic). – Johannes Kloos Nov 28 '12 at 19:00
Thank You....... – user45099 Nov 28 '12 at 19:36
True, if $I$ is false, $I \to (J \to K)$ is true (assuming the conditional is truth-functional). But so what? That fact is no help at all if the aim to establish the truth of $K$ ... – Peter Smith Nov 28 '12 at 22:21
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. – andybenji Nov 28 '12 at 22:54

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