  There are many kinds of conditional for which this inference fails. If “\(\rightarrow\)” has modal force (if it expresses a kind of entailment, in which \(p \rightarrow q\) is true when in every related circumstance in which \(p\) holds, \(q\) does too), and if “\(\vdash\)” expresses local consequence \((p\vdash q\) if and only if any model, at any circumstance at which \(p\) holds, so does \(q)\) it fails. It may be true that Greg is a logician \((p)\) and it is true that Greg’s being a logician entails Greg’s being a philosopher \((p \rightarrow q\) – in related circunstances in which Greg is a logician, he is a philosopher) but this does not entail that Greg is a philosopher. (There are many circumstances in which the entailment \((p \rightarrow q)\) is true but \(q\) is not.) So we a circumstance in which \(p\) is true but \((p \rightarrow q) \rightarrow q\) is not. The argument is invalid.