Odd proof method

Solution 1:

This is a standard method of proof called proof by cases (or proof by exhaustion). It works for any finite number of cases. Suppose you know that $P_1$ or $P_2$ or ... $P_n$ must be true, i.e. at least one of the $P_i$ is true. If you can prove that $Q$ is true in each case (assuming each of the $P_i$ in turn is true), then $Q$ must be true.

In your example, you have two cases: (1) RH is true, (2) RH is false.

Solution 2:

This proposal seems like an amazingly roundabout method of proof. The theorems that led to this question use RH (and $\neg$ RH) in important but not critical ways. For example, we need to bound the growth of some function; both RH and $\neg$ RH provide different ways to bound that growth. If we could find a way to bound the growth using neither, then we would have a proof independent of RH.

Solution 3:

"Formally" it works.

From the two proofs [in all the argument, I left implicit a "common set" of assumptions : $\Gamma$] :

$\vdash P \rightarrow Q$

and

$\vdash \lnot P \rightarrow Q$

by the tautology :

$\vdash (P \rightarrow Q) \rightarrow [(\lnot P \rightarrow Q) \rightarrow Q]$

we may have, by modus ponens twice :

$\vdash Q$.

Now, with the additional proof :

$\vdash Q \rightarrow P$

with the above result, by modus ponens, we may conclude also :

$\vdash P$.

The only doubt I have is : why, in general, we may expect taht it is "easier" to produce three proofs (of $P \rightarrow Q$, $\lnot P \rightarrow Q$ and $Q \rightarrow P$) instead of two : $\vdash Q$ and $\vdash P$ ?

Solution 4:

If (p⇔q), then (p→q), as well as (q→p). Thus, if (p⇔q), and ($\lnot$p→q), since [(p→q)→(($\lnot$p→q)→q)] is a theorem or axiom in any complete system of propositional calculus, we can obtain "q". Since we have (q→p) and "q", we can also obtain "p" by detachment. Therefore, if (p⇔q) and ($\lnot$p→q), then, yes, it comes as valid to infer both "p" and "q".

Or, in Polish notation:

If Epq, then Cpq, as well as Cqp. Thus, if Epq, and CNpq, since CCpqCCNpqq is a theorem or axiom in any complete system of propositional calculus, we can obtain "q". Since we have Cqp and q we can also obtain "p" by detachment. Therefore, if Epq and CNpq, then, yes, it comes as valid to infer both "p" and "q".