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I am having some difficulties understanding logical arguments. I was taught that the notion of a valid argument is formalized as follows:

"An argument $P_1, P_2,\cdots , P_n ⊢ Q $ is said to be valid if $Q$ is true whenever all the premises $P_1, P_2,\cdots , P_n$ are true. An argument which does not satisfy this condition is a fallacy."

But (according to my book) the argument $p \rightarrow q,\quad q\;⊢ \;p\; $ is a fallacy, but I don't see why. If we construct a truth table

$p$ | $q\;$ | $p \rightarrow q$ |

T | T | T |

T | F | F |

F | T | T |

F | F | T |

we see that when $p$ and $q$ are true, $p \rightarrow q$ is also true (line 1). So then, why isn't this valid? I do know that for an argument to be valid, with premises $P_1, P_2,\cdots , P_n$, the proposition $(P_1 \land P_2 \land \cdots \land P_n) \rightarrow Q$ should be a tautology, but that doesn't dispel my confusion.

What am I missing?

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    All valid arguments in **propositional** logic can be mechanically identified via truth tables.2011-07-07

3 Answers 3

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To expand upon what Carl Mummert said,an argument is invalid if there is an instance where all of the premises are true and the conclusion is false. As stated this happens on the third line of your truth table, as 'p' is your conclusion.

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In the third line of the truth table, $q$ and $p \to q$ are both true and $p$ is false. So $q, p\to q \vdash p$ is not a valid argument.

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    To perhaps clarify (or perhaps just belabor) Carl's response, the premises of your argument are (p->q) and q, while the conclusion is p. Thus by the third line the argument is invalid.2011-07-08
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I think one the important points is that truth-functionality is too coarse for certain applications/issues, in the sense that the inner-structure of the statements does not matter, e.g., the coarseness does not prevent statements of this sort:

All Men are Mortal:=P


Adam is not Mortal:=Q

Since we can just assign to P the value T , independent of its actual meaning, same for Q.

More formally, the issue of material implication guarantees that if Q is true in P->Q, then (P->Q) itself is true, so that:

($(P\rightarrow Q)$ /\Q) is always true when Q is. In particular, this last argument is true when Q is true and P is false. This assignment, then, of P:=T and Q is false is one which makes your premises true and your conclusion false.