How is the statement $\bot \to \top$ true? If, for example, I take the statement
$P = 2$ is an odd number
$Q = 5$ is an odd number
Then $P$ is false and $Q$ is true.
Clearly, false implies true translates to
if $2$ is an odd number then 5 is an odd number
which means that
if $2$ is not an odd number (i.e. even) then $5$ is not an odd number
which is clearly false as $2$ is indeed an even number. Please help me to find what I’m missing out in this. Thanks for your help!
On
...“if 2 is an odd number then 5 is an odd number” , which means that “if 2 is not an odd number (i.e. even) then 5 is not an odd number ...
No, that does not follow. If $p \to q$, then $\neg q \to \neg p$, rather than $\neg p \to \neg q$
So, “if 2 is an odd number then 5 is an odd number” implies that “if 5 is not an odd number then 2 is an odd number”
And that actually makes some sense: given that $5$ is an odd number, to say that $5$ is not an odd number implies contradiction, and anything follows from a contradiction. Indeed, the logic is the same as: "If that's true, then pigs fly!".
On
I think you are looking for Duns Scotus’ law, or ex falso quodlibet (https://en.m.wikipedia.org/wiki/Principle_of_explosion). In short, whenever the premise of the implication is false, the entire implication is necessarily true.
You can verify the validity of this statement directly by inspecting the truth table for implication.
In you example you appear to be making an error. You said that since $$P \rightarrow Q$$ then this means that $$\neg P \rightarrow \neg Q$$ But this does not follow. Instead, the correct conclusion is that $$\neg Q \rightarrow \neg P$$ by contraposition.
On
Your two Axioms are: P = 2 is an odd number and Q = 5 is an odd number. Taking your Axiom into account you would indeed get the answer true, because you state that P is odd and Q is odd which would yield, odd implies odd which is true. For the second example with negation it would also yield true, because not odd implies not odd which is true.
You seem to think that an implication $P\implies Q$ also implies its so-called converse, which is the statement $\text{not }P\implies \text{not }Q$. This is not true.
For instance, take the statement "if it's raining outside, then the ground is wet". This is (usually) true, while the converse statement "if it's not raining, then the ground is dry" is not true, because there are many other ways for the ground to become wet.
What an implication $P\implies Q$ does imply is the so-called contrapositive $\text{not }Q\implies \text{not }P$. The contrapositive of the above statement is "if the ground is dry, then it's not raining", which you will recognize as a (again, usually) true statement.
Also, back to your problem of comprehending why $\text{false}\implies\text{true}$ is considered true, this some times clashes with people's intuition (as it has with yours). I like to explain it by thinking of implications as promises. Thus "If 2 is an odd number, then 5 is an odd number" can be rephrased as "If 2 is an odd number, then I can promise you that 5 is an odd number". This is a promise that I can (provably) keep (or rather, no one can make me break it), and therefore the statement is considered true.
To draw that promise analogy further, "If it's raining outside, then I promise that the ground is wet" is (usually) a safe promise, while "If it's not raining outside, then I promise the ground isn't wet" is not so safe (maybe some kids played with a water hose, or the road was just pressure washed, or something).