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For $F\to T$ we are starting from a false point and concluded the result to be false and the result of $F\to F$ is true. But how someone can start at a false point and prove something correct? I am talking about $F\to T$. I know that result of $F\to T$ is true because it's not breaking the if..then promise. I am searching a practical example where someone has started with a false proposition and proved something correct?

Mr Pie
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OldSchool
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  • But, having said that, we have the so-called principle of explosion: "according to which any statement can be proven from a contradiction. That is, once a contradiction has been asserted, any proposition (including their negations) can be inferred from it". Thus, from a contradiction we can correctly infer true as well as false conclusions. – Mauro ALLEGRANZA Oct 02 '17 at 10:47
  • @MauroALLEGRANZA but the thing you wrote is a subset of meaning of P -> Q. This is the thing we often do in proof by induction where we prove P(n+1) by P(n) and write P(n) -> P(n+1). – OldSchool Oct 02 '17 at 10:48
  • See this recent post for a discussion of the inference rule. – Mauro ALLEGRANZA Oct 02 '17 at 10:49
  • Ok; thus, the "meaning" of the rule is: from a contradiction, anything follows (ex falso (sequitur) quodlibet). – Mauro ALLEGRANZA Oct 02 '17 at 10:51
  • Proof: 1) $0=1$ --- assumed; 2) from arithmetical axioms, we can prove that $0 \ne 1$; thus: 3) contradiction ! and applying the above rule: 4) $1=1$. In this way, we have proved that from $0=1$ we can prove $1=1$ and thus, as per your comment, we conclude with: $0=1 \to 1=1$. – Mauro ALLEGRANZA Oct 02 '17 at 10:54

2 Answers2

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How about this?

If $3 = -3$, then $9=9$ (obtained by squaring both sides--if you square both sides of an equation, the result will be an equation).

paw88789
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The fact the logically $P\implies Q$ is true when $P$ is false and $Q$ is true does not mean that one ever attempts to prove $Q$ from a premise $P$ known to be false. Once the implication is proved true it is always used by invoking modes ponens, that is once $P\implies Q$ is established, and $P$ is established, one deduces $Q$. So, there is no practical scenario where you would establish that $Q$ is true by first establishing that $P\implies Q$ is true and then show that $Q$ is false. Simply because there is no valid inference rule that permits that.

Ittay Weiss
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  • “So, there is no practical scenario where you would establish that Q is true by first establishing that P⟹Q is true and then show that Q is false.” There is no theoretical scenario either. “P⟹Q is true when P is false and Q is true” allows to establish P⟹Q, and you are trying to establish Q. – beroal Oct 02 '17 at 12:15
  • Yes @beroal that's what I said. – Ittay Weiss Oct 02 '17 at 12:23