Inner products equal implies the arguments are equal

Question

Today I was looking at theorems involving inner products which use the 'fact' that if $\langle x,y \rangle = \langle x,z \rangle$ for all $x$ then $y=z$. Is the following a sufficient proof of that? Consider an inner product space $(M,\langle\cdot,\cdot\rangle)$ and $x,y,z\in M$. Assume that $\forall x\in M, \langle x,y\rangle = \langle x,z\rangle$. Then, using the linearity of the inner product, $\forall x\in M, \langle x,y-z\rangle =0$, meaning that $y-z$ is orthogonal to every vector in $M$. Thus, since $0$ is the only vector which is orthogonal to all vectors, $y-z$ is $0\implies y=z$. Or alternatively, since $\forall x\in M, \langle x,y-z\rangle=0$, we can pick $x=y-z$ to get $\langle y-z,y-z\rangle = 0 =||y-z||^2\implies y-z=0\implies y=z$.

Answer 1

The first proof uses circular reasoning; it essentially shows that $0$ is the only vector that is orthogonal to all other vectors in $M$, which is an assumption in the proof. Your second proof is the way to go.