In a previous question I have been told that lambda calculs is pure syntax. I see that Lambda calculus is introduced inductively, but I don't see from what axioms it follows that: $$(\lambda x.x) M \leadsto_\beta M$$ Because in the reduction above $\lambda x.x$ is seen as a function over Lambda terms.
Can somebody explain me what is the point here?
Posted as an answer, as requested by the OP.
Writing $$(λx.\ x)\ M=M$$ is wrong (or yields falsehood), those two terms are not equal.
I think what you want is the $β$-reduction, that is, $$(λx.\ x)\ M\ {\leadsto}_\beta\ M,$$ however, as the name suggests, reduction is not equality. In case of $\beta$-reduction, the definition clearly states why $(\lambda x.\ x)\ M\ \leadsto_\beta \ M$, namely $(\lambda x.\ x)\ M$ rewrites to $x[x \mapsto M]$ which is $M$.
I hope this helps $\ddot\smile$