$\lambda$-calculus: structural induction principle over $\Lambda$

Question

The set $\Lambda$ is given inductively by:

• $x\in\Lambda$, if $x$ is a variable;
• $(\lambda x M)$, if $x$ is a variable and $M\in\Lambda$;
• $(MN)$, if both $M,N\in\Lambda$.

Now, consider the structural induction principle associated with $\Lambda$. Say a property $P(M)$ over $M\in\Lambda$. Consider just the second case; is it(?)

• $P(x)\wedge P(M)\implies P(\lambda x M)$, for all $x$ variable and $M\in\Lambda$.

Answer 1

No. $P(x)$ doesn't even make sense, because $P$ is a property of lambda expressions, and $x$ is not an expression, it is just a variable. The correct induction principle is

• $P(M) \implies P(\lambda x M)$ for all variables $x$ and $M\in\Lambda$.