I'm reading Lambda-Calculus and Combinators: An Introduction, and there's the following definition of $\lambda$-substitution:
- $FV(P)$ stands for the set containing all free-variables from $P$.
Definition 1.12 (Substitution) For any $M, N, x$, define $[N/x]M$ to be the result of substituting $N$ for every free occurrence of $x$ in $M$, and changing bound variables to avoid clashes. The precise definition is by induction on $M$, as follows (after [CF58, p.94]).
(a) $[N/x]x \equiv N$
(b) $[N/x]a \equiv a$ for all atoms $a \not \equiv x$
(c) $[N/x](PQ) \equiv ([N/x]P)([N/x]Q)$
(d) $[N/x](\lambda x.P) \equiv (\lambda x.P)$
(e) $[N/x](\lambda y.P) \equiv P$ if $x \not \in FV(P)$.
(f) $[N/x](\lambda y.P) \equiv \lambda y. [N/x]P$ if $x \in FV(P)$ and $y \not \in FV(N)$.
(g) $[N/x](\lambda y.P) \equiv \lambda z. [N/x][z/y]P$ if $x \in FV(P)$ and $y \in FV(N)$.
I do understand that:
- $(a), (b)$ are the base cases for this induction.
- $(d)$ exists per definition, as one is not allowed to substitute bound variables.
- $(g)$ prevents a bound variable from changing to a free one. It does this by first substituting a bound variable.
My question is: If one deletes $(d)$ and allows bound variables to be substituted, is $(g)$ strong enough to handle it without messing up everything?
I'm asking this because $(d)$ seems to prevent the following $\alpha$-equivalent substitutions.
$$ [y/x] ~~ \lambda x. x \equiv \lambda y.y $$