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I can not see the difference between beta reduction and substitution.

For example:

(+ x 1) [x -> 2]

Here I can do the substitution, replace the variable x through 2(lambda expression) and result is:

(+ 2 1)

An example of beta reduction:

(λ x. + x 1) 2

the result would be 

+ 2 1

For me, there are no difference between them, but there is a rule, I can not substitute the lambda expression, when the metavariable is bound

(λ x. + x 1) [x -> 2] No possible to substitute it.

Only free variables can be substitute it.

What is the difference between the diction:

(λ x. + x 1) 2

and 

(λ x. + x 1) [x -> 2]
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1 Answers1

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The $\beta$ reduction is defined by using capture-avoidng substitution. That is,

$$ (\lambda x. M)N \longrightarrow_\beta M[x \mapsto N] $$

So it's the definition. Under the $\beta$-equivalence, $(\lambda x. M)N \equiv_\beta M[x \mapsto N]$, but syntactically, $(\lambda x. M)N \neq M[x \mapsto N]$.

Note that substitution is a meta-operation on $\lambda$-calculus terms (normally). A $\lambda$-calculus term $M$ is whether a variable $x$, a lambda abstraction $(\lambda x. M')$, or an application $M_1 M_2$. That's it. There is no substitution. Substitution is a map on $\lambda$-calculus terms. Therefore, for example, a term $({+}~x~y)[x \mapsto z]$ is syntactically equal to $({+}~z~y)$. There is no reduction. In contrast to this, $(\lambda x. {+}~x~y)~z$ is not syntactically equal to $({+}~z~y)$, although it can be reduced to $({+}~x~y)$.

Also, $(\lambda x. {+}~x~1)~2$ is different from $(\lambda x. {+}~x~1)[x \mapsto 2]$. The latter is equal to $(\lambda x. {+}~x~1)$.

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nekketsuuu
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