# How do I prove that the MSE is zero when all predictions are equal to the corresponding labels?

In the back-propogation algorithm, the error term is:

$$E=\frac{1}{2}\sum_k(\hat{y}_k - y_k)^2,$$

where $$\hat{y}_k$$ is a vector of outputs from the network, $$y_k$$ is the vector of correct labels (and we work out the error by calculating predicted-observed, squaring the answer, and then summing the answers for each $$k$$ (and dividing by 2).

How do you prove that if this answer is $$0$$ (i.e., if $$E=0$$), then $$\hat{y}_k=y_k$$ for all $$k$$?

Let's first prove that, if $$\hat{y}_k = y_k$$, then the $$E = 0$$. I will leave all steps, so that it's super clear.
\begin{align} E &=\frac{1}{2}\sum_k(\hat{y}_k - y_k)^2 \\ &=\frac{1}{2}\sum_k(y_k - y_k)^2\\ &=\frac{1}{2}\sum_k(0)^2\\ &=\frac{1}{2}\sum_k 0\\ &=\frac{1}{2} 0\\ &=0\\ \end{align}
To prove the other way around, i.e. if $$E = 0$$, then $$\hat{y}_k = y_k$$, you can do as follows
\begin{align} \frac{1}{2}\sum_k(\hat{y}_k - y_k)^2 &=E\\ &=0 \end{align} Recall now that any number squared is non-negative (i.e. positive or zero). Given that $$(\hat{y}_k - y_k)^2$$ is non-negative, then $$\sum_k(\hat{y}_k - y_k)^2$$ is a sum of non-negative numbers. The only way that a sum of non-negative numbers is equal to zero is if all numbers are zero, so we must have $$\hat{y}_k = y_k$$ (because any non-zero number squared is non-zero).
(Note that $$E$$ is the mean squared error, i.e. a loss function, and it's not the back-propagation algorithm, which is just the algorithm that you use to compute partial derivatives of $$E$$ with respect to the parameters of the model, which are not even visible in the way you wrote $$E$$).