Understanding Vacuously True (Truth Table) [duplicate]
I don't know very much about formal logic, and I'm trying to understand the concept of vacuously true statements. Consider the truth table below: $$\begin{array} {|c|} \hline P & Q & P\implies Q & Q\implies P & P\iff Q \\ \hline T & T & T & T & T \\ \hline T & F & F & \color{blue} T & F\\ \hline F & T & \color{blue}T & F & F\\ \hline F & F & \color{blue} T & \color{blue}T & T \end{array}$$.
The blue letters are definitions. To see why these definitions are the correct choices (as opposed to $F$s), suppose we changed the lower left $\color{blue} T$ to an $F$ (this would then force the lower right $\color{blue}T$ to an $F$), so $P\iff Q$ would be $F$ for $P$ and $Q$ both $F$, which isn't want we want. So I see why this makes sense to choose the lower entries as $\color{blue}T$.
However, it isn't clear to me why $P\implies Q$ true for $P$ false and $Q$ true is the sensible choice (same for $Q\implies P$ true for $Q$ false and $P$ true). For if the $\color{blue}T$s in the middle rows were switched to $F$s, then $P\iff Q$ would still be $F$. So I don't see what the problem would be. Can someone please explain?
Here is why we say $P \implies Q$ is true if $P$ is false and $Q$ is true:
Let $P$ be the statement "it's raining outside" and $Q$ be the statement "the car is wet".
In order for $P \implies Q$ to be true, what needs to happen is: every time it rains outside, it better follow that the car is wet. That's all you need to check.
So the only time $P \implies Q$ is false is when we get that it's raining outside and the car isn't wet.
When it's not raining, we don't have that it's raining outside and the car is not wet. Since we don't have this, that means $P \implies Q$ isn't false, since the only time it is false is when we get that it's raining outside and the car isn't wet. If it's not raining outside, we don't get this, so the statement is not false, so it must be true.
Consider the implication "If it is raining, then it is cloudy."
$$Raining\implies Cloudy$$
In the mathematical sense of implication, this is not a claim of a causal relationship between rain and cloudiness: e.g. that cloudiness somehow causes rain, or that rain somehow causes cloudiness. This is also not a claim of any kind of historical pattern over time. An implication can be applied to a single instant in time, e.g. to logical relationships that may be evident even in a single still photograph. In the mathematical sense of implication, it is only being claimed that, at a given instant in time, it is not both raining and not cloudy.
$$Raining \implies Cloudy \equiv \neg[Raining \land \neg Cloudy]$$
There is no notion of cause and effect in mathematics. There is no passage of time. These are the realms of science. In mathematics, time is just another variable of no particular significance.
$$\begin{array}{c|c|c|c} \space&Raining&Cloudy&Raining\implies Cloudy\\\hline 1&T&T&T\\ 2&T&F&F\\ 3& F&T&T\\ 4&F&F&T \end{array}$$
From this truth table, we see that, at any given instant in time...
- If $Raining\implies Cloudy$ is true and $Raining$ is true (line 1), then $Cloudy$ will be true.
- If $Raining \implies Cloudy$ is true and $Cloudy$ is false (line 4), then $Raining$ is false.
- If $Raining \implies Cloudy$ is false (line 2) then $Raining$ is true and $Cloudy$ is false
- If $Raining$ is false (lines 3, 4), then $Raining \implies Cloudy$ will be true
- If $Cloudy$ is true (lines 1, 3) , then $Raining \implies Cloudy$ will be true
Follow-up: The truth table for material implication does not depend solely on the above definition. See my answer at How does one know if $A \implies B$ (an implication) is true without knowing if $B$ (the consequent is true) is true?
$P \Rightarrow Q$ is logically equivalent to $\sim P \vee Q$ (where $\sim$ is '"not" and $\vee$ is "or"), and $\sim P \vee Q$ is true whenever $P$ is false. Another thing to think about is providing counterexamples. $P\Rightarrow Q$ can only be proved false if you give an example where $P$ is true and $Q$ is not true. Any example where $P$ is false is not a counterexample, so $P\Rightarrow Q$ is vacuously true in that case.