Each variable is an expression. For any expressions M and N and variable a, the following are also expressions:Variables:`x`

,`y`

,`z`

...

We'll tend to writeAbstract:`(λa M)`

`(λa M)`

as just `(\a M)`, so we don't have to write out the markup code for the `λ`

. You can yourself write `(λa M)`

or `(\a M)` or `(lambda a M)`.
Some authors reserve the term "term" for just variables and abstracts. We'll probably just say "term" and "expression" indiscriminately for expressions of any of these three forms. Examples of expressions: x (y x) (x x) (\x y) (\x x) (\x (\y x)) (x (\x x)) ((\x (x x)) (\x (x x))) The lambda calculus has an associated proof theory. For now, we can regard the proof theory as having just one rule, called the rule of **beta-reduction** or "beta-contraction". Suppose you have some expression of the form: ((\ a M) N) that is, an application of an abstract to some other expression. This compound form is called a **redex**, meaning it's a "beta-reducible expression." `(\a M)` is called the **head** of the redex; `N` is called the **argument**, and `M` is called the **body**. The rule of beta-reduction permits a transition from that expression to the following: M [a:=N] What this means is just `M`, with any *free occurrences* inside `M` of the variable `a` replaced with the term `N`. What is a free occurrence? > An occurrence of a variable `a` is **bound** in T if T has the form `(\a N)`. > If T has the form `(M N)`, any occurrences of `a` that are bound in `M` are also bound in T, and so too any occurrences of `a` that are bound in `N`. > An occurrence of a variable is **free** if it's not bound. For instance: > T is defined to be `(x (\x (\y (x (y z)))))` The first occurrence of `x` in T is free. The `\x` we won't regard as being an occurrence of `x`. The next occurrence of `x` occurs within a form that begins with `\x`, so it is bound as well. The occurrence of `y` is bound; and the occurrence of `z` is free. Here's an example of beta-reduction: ((\x (y x)) z) beta-reduces to: (y z) We'll write that like this: ((\x (y x)) z) ~~> (y z) Different authors use different notations. Some authors use the term "contraction" for a single reduction step, and reserve the term "reduction" for the reflexive transitive closure of that, that is, for zero or more reduction steps. Informally, it seems easiest to us to say "reduction" for one or more reduction steps. So when we write: M ~~> N We'll mean that you can get from M to N by one or more reduction steps. Hankin uses the symbolApplication:`(M N)`

`→`

for one-step contraction, and the symbol `↠`

for zero-or-more step reduction. Hindley and Seldin use `⊳`_{1}

and `⊳`

.
When M and N are such that there's some P that M reduces to by zero or more steps, and that N also reduces to by zero or more steps, then we say that M and N are **beta-convertible**. We'll write that like this:
M <~~> N
This is what plays the role of equality in the lambda calculus. Hankin uses the symbol `=` for this. So too do Hindley and Seldin. Personally, I keep confusing that with the relation to be described next, so let's use this notation instead. Note that `M <~~> N` doesn't mean that each of `M` and `N` are reducible to each other; that only holds when `M` and `N` are the same expression. (Or, with our convention of only saying "reducible" for one or more reduction steps, it never holds.)
In the metatheory, it's also sometimes useful to talk about formulas that are syntactically equivalent *before any reductions take place*. Hankin uses the symbol `≡`

for this. So too do Hindley and Seldin. We'll use that too, and will avoid using `=` when discussing metatheory for the lambda calculus. Instead we'll use `<~~>` as we said above. When we want to introduce a stipulative definition, we'll write it out longhand, as in:
> T is defined to be `(M N)`.
We'll regard the following two expressions:
(\x (x y))
(\z (z y))
as syntactically equivalent, since they only involve a typographic change of a bound variable. Read Hankin section 2.3 for discussion of different attitudes one can take about this.
Note that neither of those expressions are identical to:
(\x (x w))
because here it's a free variable that's been changed. Nor are they identical to:
(\y (y y))
because here the second occurrence of `y` is no longer free.
There is plenty of discussion of this, and the fine points of how substitution works, in Hankin and in various of the tutorials we've linked to about the lambda calculus. We expect you have a good intuitive understanding of what to do already, though, even if you're not able to articulate it rigorously.