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In propositional logic, for the contradiction rule:

$\neg P \rightarrow Q \text{ (Q is a contradiction)}$

$\therefore P$

Why could $\neg P \rightarrow Q$ be possible if $\neg P$ is false? Because P and Q are independent, only when we firstly assume $\neg P$ is true can we imply a contradiction. But actually, $\neg P$ is false and how can we conclude a contradiction when $\neg P$ is false?

For example, the precondition is x = 2. Let P be x is an even number. The presumption $\neg P$ that x is not an even number would lead to a contradiction. But actually this would not happen because Q can only happen when $\neg P$ is true. However, the precondition x = 2 makes Q never happen.

The proof of the contradiction rule in the textbook says because we need to make the condition statement here true to infer the conclusion P to be true, thus due to the contradiction Q, $\neg P$ must be false. But the truth of P or Q only have something to do with the real situation, which means Q cannot hold, so why does the proof make sense?

Steven Shen
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  • An implication with a false premise is always true. The only time an implication (the entire proposition $a\to b$ is false, is when a is true, but b is false. – amWhy Jan 01 '23 at 11:52
  • @amWhy That's true. But then how can we find such Q? – Steven Shen Jan 01 '23 at 11:56
  • We assume that not-P is true. This produce a contra and tus we reject the assumption. If the assumption that not-P is true is unteneable, then... – Mauro ALLEGRANZA Jan 01 '23 at 14:24

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Allow me to change your example a bit: Suppose the precondition is that $x$ is an even prime number greater than $2$. Now, note that that clearly implies a number of things: it implies that $x$ is even, it implies that $x$ is a prime number, and it implies that $x$ is greater than $2$. But of course there are no even prime number greater than $2$: the precondition is false! And yet, it also clearly implies lots of things.

So: false statements can imply other statements. Even statements that can never be true still imply things. The statement $P \land \neg P$ clearly implies $P$, and also clearly implies $\neg P$. So even contradictions imply things. (In fact, as it turns out, contradictions imply any statement! )

Bram28
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But actually this would not happen because $Q$ can only happen when $\neg P$ is true.

No, $\neg P \to Q$ does not mean that $Q$ only happens when $P$ is false. It means that if $P$ is false, then $Q$ happens.

Patrick Stevens
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Fundamentally, proposition P can be true or false (assuming traditional Boolean logic). We do not need to assign a truth value to P in order to perform calculations with P. If we knew the truth value, we would not need a variable.

So, we do not know the truth of P (for now).

However we discover that the logical implication of Not P implies Q is a contradiction. The only valid conclusion is that Not P was False. Since we are working in traditional Boolean logic, P can only have two states.

If Not P is false then, by definition, P must be True.

Now, if we look back over our work, P must have been true all along. We just didn't know it at the time.

It is a bit (but only a bit) like writing $2x+1=5$ and then concluding $x=2$.

The value of $x$ must have been 2 all along, we just didn't know it until we solved the equation.

Red Five
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