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Could you help me read (interpret) the truth tables of the two operators?

For the implication operator, the truth table is:

$$\begin{array}{c c | c} h& c& h \implies c \\ \hline T& T& T& \\ T& F& F& \\ F& T& T& \\ F& F& T& \\ \end{array}$$

Is the third column indicating when the operator holds, in other words, when a value of $s$ CAN imply a value in $t$?

For the "if and only if" conditional connective, the truth table is:

$$\begin{array}{c c | c} s& t& s \iff t \\ \hline T& T& T& \\ T& F& F& \\ F& T& F& \\ F& F& T& \\ \end{array}$$

Ziezi
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2 Answers2

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Formally, $p \implies q$ means "(not $p$) or $q$." Thus if $p$ is false then not $p$ is true, so "(not $p$) or $q$" is also true.

Informally, $p \implies q$ means that if you know $p$ is true then you can conclude that $q$ is true. So if $p$ is false, then $p \implies q$ does not say anything about $q$. So the implication is always true. Whatever one wants is implied by a false statement.

quid
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  • Thank you for the great examples! Is the truth table saying that there are three ways in which you can prove $q$ from $p$? – Ziezi Sep 18 '16 at 17:27
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    You are welcome. No it does not say this. It just explains what it means that "$p \implies q$." You can derive $q$ from $p$ only if you know that $p \implies q$ is true and you know that $p$ is true. – quid Sep 18 '16 at 17:32
  • OK, if I've understood well, then the last two lines together (of the truth table) are equivalent to your last sentence, namely: "Whatever one wants is implied by a false statement." – Ziezi Sep 18 '16 at 17:43
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    Yes, this a way to paraphrase it. – quid Sep 18 '16 at 17:49
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The third column is showing you when the implication is false, an example is $$p \implies q$$ If p is true and q is false, the implication is false. "since truths cannot reach lies".
The "if and only if"(iff) is a doble implication, $p \iff q$ is equivalent to say $p \implies q \land q \implies p$

sango
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