Explain Godel's Second Incompleteness Theorem: Why is A Theory Unable to Prove its Consistency if it is Consistent?

Question

From Peter Smith's Godel book (p. 6)

"ConT" means "theory T is consistent"

"GT" is "Godel unprovable sentence"

ConT $\to$ GT inside T

...it immediately follows that if T is consistent it cannot prove ContT.

Can someone please explain? Many thanks.

Answer 1

Let $T$ be a consistent theory. Then, by definition of consistency, for every sentence $\rho$ in $T$, it's not the case that both $\rho$ and $\neg \rho$ are provable in $T$.

As you stated, if $T$ is consistent, then $T$ contains at least one Godel sentence. Thus, there is at least one Godel sentence in $T$. By definition, a Godel sentence $\phi$ of $T$ is a well-formed formula with the following characteristics:

$1$. $\phi$ is constructed in such a way that it effectively means "This sentence is not provable in $T$."

$2$. $\phi$ is in fact true.

Hereafter, lets assume $T$ can prove its own consistency. This implies that for every sentence $\rho$ in $T$, $T$ can actually prove it's not the case that both $\rho$ and $\neg \rho$ are provable in $T$. In other words, for every sentence $\rho$, $T$ can prove that $\rho$ is not provable or that $\neg \rho$ is not provable in $T$. Well, if $T$ can prove this about every sentence in $T$, then it can certainly prove this about a Godel sentence $\phi$ in $T$. Here is where things get interesting...

Suppose $T$ can prove $\phi$ is not provable in $T$. This immediately leads to a contradiction because, as a result of $1$ and $2$ above, it cannot be proven in $T$ that $\phi$ is not provable in $T$. Now suppose $T$ can prove $\neg \phi$ is not provable in $T$. Again, we arrive at a contradiction because $\neg \phi$ effectively means "This sentence is provable in $T$," implying $T$ will have proven that a provable statement is not provable.

Since each case above results in a contradiction under the assumption that $T$ can prove its own consistency, this assumption must be false. In other words, it must be the case $T$ cannot prove its own consistency.

Therefore, if $T$ is a consistent theory, then $T$ cannot prove its own consistency.