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Let's say I attempted to solve a logical statement in the form using contradiction:

$\forall x \in \Bbb R, (P \implies Q)$


$\exists x \in \Bbb R, (P \land \lnot$ Q).

Initially I did not know if the original statement was true or false. And I proved that the negation was true and the original statement was false Is this still a proof by contradiction or just a proof by negation? Im just confused because the question came from the contradiction chapter of the book... Because I did not contradict any of my new assumptions I made when I negated the proof.

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A proof by contradiction would be if assume what you negate Q and you also assume P. And show that the negated Q cannot be true if I assume P. Basically trying to find a mistake that contradicts your assumption. – user60887 Nov 13 '13 at 4:47
Thats what I attempted to do, but the assumption ended up being true and i could not find a contradiction. – Matt Nov 13 '13 at 5:01
what were you trying to prove at first? – user60887 Nov 13 '13 at 5:44
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There is a good thread on that here: Proof by contradiction vs Prove the contrapositive

You have proved the negation of your premise, which is slightly different in that you did not get a contradiction in the actual steps of your proof, per se. However, the assumption was that your original staement was true, hence the negation should be false. Since you found the negation true, it cannot be the case that your original statement is also true, so assuming your original is true leads to a contradiction (P and not P are both true). Perhaps that is where the contradiction lies?

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Proof by contradiction is a method of deduction which means assuming some proposition and find a contradiction out of your assumption. This would lead you that your assumption was false, then its opposite is true.

To prove the contraposition is not a method of proving things, it's only a shortcut which can only be used for if...then statements. It is just proving $\neg Q \implies \neg P$ instead of $P\implies Q$. However for prove $P \implies Q$ by contradiction, you have to show that $P$ is true and assume $Q$ is false and find a contradiction. You can even use proof by contradiction to just one proposition, $P$.

In natural deduction, prove by contrapositive cannot be used as it is not a logical rule but a result of the truth table of "if...then" statements. You cannot just switch $P \implies Q$ into $\neg Q \implies \neg P$ while making a natural deduction. But you can always say "Assume $\neg P$, there is a contradiction, then $P$".

Another addition: $\forall x \in \Bbb R, P\implies Q$ is not a good structure. It is a mixture of propositional logic and first order logic. $P$ and $Q$ does not deal with the values of $x$ there, as they are not dependent to $x$. If what you're trying to state is not something like "for any real number $x$, if tomorrow is rainy, then the traffic is gonna be a mess" but "for any real number $x$, if $x$ is odd, then $x$ is a natural number", you should say $\forall x \in \Bbb R, (P(x)\implies Q(x))$. Easiest way to prove this wrong is contradiction with counter example like:

  1. $\neg Q(-1)$ (premise)

  2. _$\forall x \in \Bbb R, (P(x)\implies Q(x))$ (assumption)

  3. $-1 \in \Bbb R$

  4. $P(-1)$

  5. $Q(-1)$($\implies$ introduction 1,2)

  6. $\bot$

  7. $\neg \forall x \in \Bbb R, (P(x)\implies Q(x))$ ($\neg$ introduction 2,6)

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