You are asked to evaluate a proof attempt of a proposition p which begins with the assumption:

Question

You are asked to evaluate a proof attempt of a proposition p which begins with the assumption: Suppose if p is false and performs a set of correct derivations and ends with the conclusion: Therefore, we conclude that p is true. Does this attempt prove the proposition p? What is(are) the propositional law(s) that you used to arrive at your decision?

Provide some context please. What is your understanding of what a mathematical proof is? What proof techniques do you know and are any of them being used (correctly or otherwise) in this proof attempt? — 79037662, Sep 20 '19 at 20:04
So you basically being asked of the proof that if $\lnot P \to P$ is demonstrated to be true, does that mean the conclusion that $P$ is true is valid? What do you think? — fleablood, Sep 20 '19 at 20:17

hamam_Abdallah · Accepted Answer · 2019-09-20T20:30:18.733

If the premisses of a valid argument are true Then you can be sure that the conclusion of your argument is also true.

But,

If the Conclusion of a valid argument is False or a contradiction, Then you can be sure that your hypothesis which is (not p) in your case is False.

In other words, $$\Bigl(\text{not} \;p \implies (q \text{ and not } q)\Bigr) \implies p$$

is a logically valid argument.

Some other valid arguments are : Modus Ponens, Modus Tollens, Disjonctive Syllogism, De Morgan's Laws,....

score 0 · Answer 2 · answered Sep 20 '19 at 20:40

You're asking whether it's true that $ (\lnot p \Rightarrow p)$ proves $p.$ The answer is yes, because if $\lnot p \Rightarrow p$, then because also $\lnot p \Rightarrow \lnot p$, it follows that $\lnot p \Rightarrow (p \land \lnot p)$, which proves $\lnot (\lnot p)$, which proves $p$.

score 0 · Answer 3 · answered Sep 20 '19 at 21:11

If your proof system is based on natural deduction, then what you are given is that $\lnot P \vdash P$. However, also by the assumption rule, $\lnot P \vdash \lnot P$. Therefore, by the negation elimination rule $\lnot E$, we get $\lnot P \vdash \bot$. But precisely by the rule known as $IP$ in classical natural deduction (for example in the listing on the right hand side at https://proofs.openlogicproject.org/), this implies that $\vdash P$.

In terms of a Fitch-like formulation of natural deduction, the resulting proof outline would look something like this:

You are asked to evaluate a proof attempt of a proposition p which begins with the assumption:

3 Answers3