Is a proof based upon $P \vee \neg P$ valid if $P$ is undecidable? In https://home.sandiego.edu/~shulman/papers/sdg-pizza-seminar.pdf, the author mentions how some proofs use the fact that $P \vee \neg P$ and then have two proofs for the $P$ is true and $P$ is false cases (the example given is the Riemann Hypothesis as $P$). He mentioned that the proofs would be invalid if the Riemann hypothesis were undecidable on page 4. The way I learned about undecidable statements was that they are true/false, but their truth/falsehood is unprovable. Is what the author said true (that the proofs are incorrect) and is my intuition on undecidable statements true as well?
 A: This Question (& the Article) is about the Philosophy of Mathematics.
In general, there are 2 types of Mathematics (or 2 types of Mathematicians) namely :  "Non-Constructive/Classical Mathematics" ("Non-Constructivists")
Who agree with the Axiom of Choice, the Principal of Excluded Middle, & such.
Who agree with statements like "There Exist X such that ..." without knowing specific values of X or Constructing such X.
"Constructive Mathematics" ("Constructivists") :
Who reject the Axiom of Choice, the Principal of Excluded Middle, & such.
Who reject statements like "There Exist X such that ..." until knowing specific values of X or until successfully Constructing such X.
The Main Issue is with "Constructing" given Items.
Example 0 (my own, not in the given article) :
Classical : We can assume Artificial Intelligence Exists, and we only have to make such Machines. We can assume Aliens Exist , and we only have to search for them.
Alternatively, We can assume there is no Artificial Intelligence, and we should not try to make such Machines. We can assume there are no Aliens , and we should not search for them. Either way, we can use these facts in our Proofs via $P \vee \neg P$.
Constructive : Until we see (or Construct) one machine with Artificial Intelligence or we see one Alien, we should not use these facts in our Proofs via $P \vee \neg P$.
The Article gives atleast 2 Examples like this :
Example 1 given in the article : Is there at least one pair of irrational numbers $a,b$ such that $c=a^b$ is rational ?
Classical : Consider $(\sqrt{2}^\sqrt{2})^\sqrt{2} = 2$ : Either $a=\sqrt{2} , b=\sqrt{2}$ makes $c$ rational OR $a=(\sqrt{2}^\sqrt{2}) , b=\sqrt{2}$ makes $c$ rational.
Constructive : NO, that is not valid ! We have to show specific $a,b,c$ to Prove the claim.
Example 2 given in the article : The topic of OP Question : Can we Prove Theorem X (unspecified in the article) ?
Classical : Either (Riemann hypothesis is true and we can show X is true) OR (Riemann hypothesis is false and we can show X is true) ; Either way X is true.
Constructive : NO, that is not valid ! We have to show whether Riemann hypothesis is true or false to Prove the claim. It might turn out that Riemann hypothesis is undecidable [[ loosely speaking, Neither true Nor false ]] , then this Proof is wrong ! Proving X by giving Examples without Riemann hypothesis will be Constructive !
Highlight: Classically, this is a valid Proof. Constructively, this is not a valid Proof.
This Question (& the Article) is about the Philosophy of Mathematics.
A: I think the message is that some people would accept such a proof using law of excluded middle, some would not. The statement that "there exists $a$, $b$ irrational such that $a^b$ is rational" has a proof using law of excluded middle that is not constructive:
https://en.wikipedia.org/wiki/Law_of_excluded_middle#Examples
Here, the statement used for excluded middle is that a number is either rational or irrational. And you can be satisfied with such a proof that shows that either cases are true that $\sqrt2^{\sqrt2}$ is irrational or rational, we can find $a$, $b$ irrational such that $a^b$ is rational.
A constructivist would demand a constructive proof, and not except this argumentation. And why not? Imagine now instead of the statement that a real number is either rational or irrational (which is true by definition), we use that either CH or not CH, even though it's undecidable. You can maybe see why some people feel uncomfortable with this. Never the less, such a proof is not correct or incorrect, it is a question of taste (or school).
We need to agree on a set of rules that makes proofs admissible. Such a set of inference rules are used in "natural deduction" to build proofs (https://en.wikipedia.org/wiki/Natural_deduction). Some systems doesn't allow using the law of excluded middle, but they have a list of deductions that are admissible. You can for example allow to get $B$ from $\neg \neg B$. If you replace this rule by the law of excluded middle, then you get another deduction system.
