First of all, I only learn "logic" in Real Analysis. So please bear with me and be welcome to define "logic" if necessary.
In First-order logic, "If P then Q" can be expressed as "(not P) or Q". Then the Truth table will tell us that if statement P is False, then the whole statement will be True regardless whether Q is true or not. I understand this, however, when it comes to the "logic" behind the implication. It confuses me.
So my question is
The assumption P is False, but we Suppose it is TRUE (if part). Even the logic behind the implication is wrong, the statement "If P then Q" is still true?
I need to emphasise that we want to assume P is TRUE though we know it is FALSE
For example, back to the time when aircraft cannot fly. Someone said, "If the aircraft can fly, then people inside it would die due to the high speed when it takes off. "
Using First-order logic, since the aircraft at that time cannot fly, so the whle statement (if-then) will be true. However, the "logic" in this statement seems to be wrong. We admit that the aircraft at that time cannot fly at that time, but we want to consider the situation where it can (maybe in the future or even just a physics thought experiment). Under this assumption, people don't die due to high speed of the aircraft. Then we would conclude that this is a False statement.
Then it will arise a natural questions like,
what should we do if we want to make inference or implication under some conditions which might not be true? Does the "logic" just lost the control after a False assumption?
How should we make a statement to distinguish "True because the logic is true" and "True because the assumption is False"
Let me rephrase my question:
Is there any theory focusing on the logic in implication instead of just true and false of the whole statement?
so that even the assumption is false, if the logic of implication does not hold then we will consider the proof/implication if False.
For example, a question on exam paper,
- "Prove $\sqrt 2 \notin \mathbf{Q}$".
If a student answers, "suppose $\sqrt 2 \in \mathbf{Q}$, then $\pi \in \mathbf{Q}$. Contradiction."
This answer technically is a True statement and indeed we arrived in a contradiction. But $\pi \in \mathbf{Q}$ does not follow from $\sqrt 2 \in \mathbf{Q}$ in logical sense. So it deserves 0 mark.
An example that is built on a false antecedent but implication is valid:
- If (1+1=1, 2+2=2^2, 3+3=3^2, etc), then n+n=n^2.
The antecedent is clearly False, but suppose instead it is true, say we redefine + as x, then logic follows. While for "if … then n+n=0", the implication doesn't follow any "logic".
Best Answer
If $P$ then $Q$ means if $P$ is true, then, no matter what, $Q$ is true.
Hence you cannot have $P$ true and $Q$ false.
So either $Q$ is true or, if $Q$ is false, since you cannot have $P$ true then $P$ is false, which is $\neg P$ true. Which you can rewrite as $$\neg P \vee Q $$
If the lights are red then cars stop. Whatever the cars are doing when the lights are NOT red will not change the truth of this statement.