Goldbach implies Bertrand

Fallacy
Goldbach Conjecture is a logical consequence of the truth of Bertrand's Postulate:


 * Bertrand's Postulate: $\forall n \in \N: n > 1:$ between $n$ and $2n$ lies a prime number $p$:


 * $n < p < 2n$


 * Strong Goldbach Conjecture: Every even number greater than $2$ can be written as a sum of two prime numbers.

Supposed Proof
Suppose that Bertrand's Postulate were wrong.

Then there exists $n \in \N$ such that:
 * $\exists p_1, p_2 \in \Bbb P: p_1 < n < 2n < p_2$

where:
 * $\Bbb P$ denotes the set of prime numbers
 * $p_1$ and $p_2$ are consecutive prime numbers, that is:
 * $\neg \exists p \in \Bbb P: p_1 < p < p_2$

Now consider the even number $2n$.

The largest prime smaller than $2n$ is less than $n$.

So it is impossible for two primes less than $2n$ to add up to $2n$.

Thus Goldbach Conjecture can not be satisfied for $2n$.

But Bertrand's Postulate, which is now called the Bertrand-Chebyshev Theorem and has been proved, states that between $n$ and $2n$ there must be a prime number.

So $p_1$ and $p_2$ can not be consecutive.

From this it follows that the Goldbach Conjecture must be true.

Resolution
Let $A$ be the statement:
 * Bertrand's Postulate holds.

Let $B$ be the statement:
 * The Goldbach Conjecture is true.

The above proof has demonstrated that:


 * $\neg A \implies \neg B$

It is supposed, then, that:
 * $A \implies B$

This is a classic instance of Denying the Antecedent.