Dubious proof of "if Goldbach’s conjecture is unprovable, it must be true"

Question

Recently, I read a news article written by an "emeritus professor at UCD school of mathematics and statistics" that made the following curious claim about the Goldbach's conjecture: "But let us suppose the conjecture is unprovable. Then it must be true." Another document by the same author reproduces the claim. But is this actually true? The article contains a proof of this statement, but I do not understand it. Actually, the proof seems invalid to me. The proof from the article is:

Because, if it were false, there would be some finite even number that is not the sum of two primes. A finite search could confirm this, making the conjecture “provably false”! In other words, falsehood of the conjecture is incompatible with unprovability. This contradiction forces us to an ineluctable conclusion: if Goldbach’s conjecture is unprovable, it must be true!

So if $G$ is Goldbach's conjecture, and $P(G)$ means that Goldbach's conjecture is provable, then we could say the first two sentences from the quote prove that:

$$\lnot G \implies P(\lnot G) \tag{1}$$

The next sentence in the quotation is rather vague. (It is unclear what the word "unprovability" refers to.) It could mean both:

$$\lnot(\lnot G \land \lnot P(G)) \tag{2.1}$$ $$\lnot(\lnot G \land \lnot P(\lnot G)) \tag{2.2}$$

Proposition $(2.1)$ is clearly untrue: Goldbach's conjecture could be both false and unprovable. In contrast, $(2.2)$ could be derived from $(1)$:

$$ \begin{align} \lnot G \implies P(\lnot G) &&\text{assuming (1)} \\ G \lor P(\lnot G) &&\text{material implication} \\ \lnot(\lnot G \land \lnot P(\lnot G)) &&\text{De Morgan laws} \end{align} $$

Considering this, I assume that $(2.2)$ is the correct interpretation of the third sentence from the quote. The last sentence of the quote puzzles me; I do not see how "$\lnot P(G) \implies G$" follows from $(2.2)$. In fact, it seems the author is working with $(2.1)$. If we were to assume the (clearly false) proposition $(2.1)$, we could indeed prove the titular claim: $$ \begin{align} \lnot(\lnot G \land \lnot P(G)) &&\text{assuming (2.1)} \tag{I}\\ \lnot P(G) &&\text{assuming the antecedent} \tag{II} \\ \lnot(\lnot G \land \top) &&\text{from (I) and (II)} \tag{III}\\ G &&\text{simplifying (III)} \tag{IV}\\ \lnot P(G) \implies G &&\text{implication introduction from (II) into (IV)} \tag{V} \end{align} $$

It seems that the author himself got confused with the word "unprovability" in the third sentence of the quotation. But is it so? Is not there any mistake in my thoughts? The author is seemingly highly educated, while I lack any tertiary education in mathematics. (I apologize for any nonstandard notation or terminology in this question.)

Answer 1

The passage makes sense if we assume the author is using "unprovability" to refer to undecidability, i.e. the situation where $G$ is neither provably true nor provably false. Defining $U\equiv \lnot P(G)\land \lnot P(\lnot G)$, we can deduce $U\implies G$ from $\lnot G\implies P(\lnot G)$.

Answer 2

As the author says it, the claim is very misleading. Something similar that is true is:

If the Goldbach conjecture is not refutable in Peano Arithmetic, then it is true.

Because suppose that the Goldbach conjecture is false. Then there is large even $n \in \mathbb N$ that cannot be written as the sum of two primes. Thus we can write the following true statement:

$\forall p < n,\ p$ is not a prime or $n-p$ is not a prime.

Here $n$ is of course represented by a numeral $n = SS\cdots S0$. The statement above can get expanded into one in $\mathsf{PA}$ that still uses only bounded quantifiers, so there is a proof of it by "brute force" and it can be written in $\mathsf{PA}$. Thus Goldbach is refutable.