Lec 1 \(\lambda\)-Calculus

Lambda calculus is a formal system in mathematical logic for expressing computation based on function abstraction and application using variable binding and substitution.

Definition

• Variables: \(x, y, z, \ldots\)
• Abstraction: \(\lambda x.M\) where \(x\) is a variable and \(M\) is a lambda term.
• Application: if \(M\) and \(N\) are lambda terms, then \((M N)\) is a lambda term.

Example

• Identity function: \(\lambda x.x\)
• Constant function: \(\lambda x.y\)
• \(\lambda x.(\lambda y.y x)\) is a function that takes two arguments and returns the second one. -- Abstraction (Definition)
• Example: \((\lambda x.(\lambda y.y x)) a b \rightarrow (\lambda y.y a) b \rightarrow b a\)
• \((\lambda x.x)y\) is a function that takes one argument and returns the argument. -- Application
• Example: \((\lambda x.x)y \rightarrow y\)
• \((\lambda x.x) (\lambda y.y x)\) is a function that takes one argument and returns the identity function. -- Application
• Example: \((\lambda x.x) (\lambda y.y x) \rightarrow \lambda y.y x\)

Free and Bound Variables

• A variable \(x\) is free in \(M\) if \(x\) is not bound by any \(\lambda\) in \(M\).
• A variable \(x\) is bound in \(M\) if \(x\) is bound by some \(\lambda\) in \(M\).

Example

• In \(\lambda x.x\), \(x\) is bound.
• In \(\lambda x.y\), \(y\) is free.
• In \(\lambda x.y x\), \(x\) is bound and \(y\) is free.
• In \((\lambda y,y)(\lambda x.xy)\), represented as \(MN\) where \(M = (\lambda y.y)\) and \(N = (\lambda x.xy)\), \(x\) is bound in \(N\) and \(y\) is bound in \(M\).
• In M: \(y\) is bound.
• In N: \(x\) is bound and \(y\) is free. also \(y\) has nothing to do with the \(y\) in \(M\).
• In \(\lambda x.(\lambda y.y xz)\), \(x\) is bound by the first \(\lambda\) and \(y\) is bound by the second \(\lambda\) and \(z\) is free.

Free Variable

CANNOT CHANGE NAME OF FREE VARIABLE

• \(FV(x) = \{x\}\)
• \(FV(\lambda x.t') = FV(t') - \{x\}\)
• \(FV(t_1 t_2) = FV(t_1) \cup FV(t_2)\)

Bound Variable

CAN CHANGE NAME OF BOUND VARIABLE * \(BV(x) = \emptyset\) * \(BV(\lambda x.t') = BV(t') \cup \{x\}\) * \(BV(t_1 t_2) = BV(t_1) \cup BV(t_2)\)

Example2

• \(BV(\lambda x.y x) = \{x\}\) ; \(FV(\lambda x.y x) = \{y\}\)
• \(BV(xyz) = \emptyset\) ; \(FV(xyz) = \{x, y, z\}\)
• \(BV(\lambda x.\lambda y.x) = \{x,y\}\) ; \(FV(\lambda x.\lambda y.x) = \emptyset\)
• \(BV(\lambda x.\lambda y.y)=\{x,y\}\) ; \(FV(\lambda x.\lambda y.y) = \emptyset\)

Closed Term

• A term with no free variables is called a closed term: \(FV(t) = \emptyset\)
• Term \(t\) 相对于 term \(t'\)封闭 if and only if \(FV(t) \cap BV(t') = \emptyset\)
• \(t\) 's free variables are not bound by \(t'\)
• Closed term s is a closed term relative to any term \(t\).

Substitution

• \([N/x]M\) implies that all occurrences of \(x\) in \(M\) are replaced by \(N\).
• In M, there cannot be any bound variable with the same name as the free variable in N.

\(\alpha\)-Renaming

• \(\forall x\) and \(\forall term\ t'\) \(\lambda x.t' =_{\alpha} \lambda y.[y/x]t'\).
• If \(t' =_{\alpha} t''\), then \(\lambda x.t' =_{\alpha} \lambda x.t''\).
• If \(t_1 =_{\alpha} t_1'\) and \(t_2 =_{\alpha} t_2'\), then \(t_1 t_2 =_{\alpha} t_1' t_2'\).

\(\beta\)-Reduction

• \(\forall x\) and \(\forall term\ t_1'\) and \(t_2\), if \(t_2\) is closed relative to \(t_1'\), then \((\lambda x.t_1')t_2 =_{\beta} [t_2/x]t_1'\).
• If \(t_1 =_{\beta} t_1'\) and \(t_2 =_{\beta} t_2'\), then \(t_1 t_2 =_{\beta} t_1' t_2'\).
• If \(t_1 =_{\beta} t_1'\), then \(\lambda x.t_1 =_{\beta} \lambda x.t_1'\).

\(\eta\)-Conversion

• \(\forall x\) and \(\forall term\ t'\), \(\lambda y.(\lambda x.t')y =_{\eta} \lambda x.t'\)
• if \(t_1 =_{\eta} t_1'\) and \(t_2 =_{\eta} t_2'\), then \(t_1 t_2 =_{\eta} t_1' t_2'\).
• if \(t_1 =_{\eta} t_1'\), then \(\lambda x.t_1 =_{\eta} \lambda x.t_1'\).

EquationSemantics

Function Composition

• composition operator: \(\circ\)
• \((f \circ g )(x) = f(g(x))\)
• \(f \circ g = \lambda x.f(g(x))\)
• \(\circ\) itself is a function : \(\lambda f.\lambda g.\lambda x.f(g(x))\)

Application

Representing Boolean Values

True and False

True: \(\lambda x.\lambda y.x\)

False: \(\lambda x.\lambda y.y\)

Example: (in steps):

• \((\lambda x.\lambda y.x) a b \rightarrow (\lambda y.a) b \rightarrow a\)
• \((\lambda x.\lambda y.y) a b \rightarrow (\lambda y.y) b \rightarrow b\)

not

• \(\lambda b.b\ (\lambda x.\lambda y.y)\ (\lambda x.\lambda y.x)\)

and

• \(\lambda a. \lambda b. a\ b\ a\)
• and True False :

\[ \begin{align*} &(\lambda a. \lambda b. a\ b\ a)\ (\lambda x.\lambda y.x)\ (\lambda x.\lambda y.y)\\ &\rightarrow (\lambda b.(\lambda x.\lambda y.x)\ b\ (\lambda x.\lambda y.x))\ (\lambda x.\lambda y.y)\\ &\rightarrow (\lambda x.\lambda y.x)\ (\lambda x.\lambda y.y)\ (\lambda x.\lambda y.x) note :\ True\ chooses\ the\ first\ one \\ &\rightarrow(\lambda x.\lambda y.y)\\ \end{align*} \]

or

• \(\lambda a. \lambda b. a\ a\ b\)

xor

• \(\lambda a. \lambda b. a\ (not\ b)\ b\)

Representing Natural Numbers

\[ \begin{align*} &\bar{0} = \lambda s.\lambda z.z =z\\ &\bar{1} = \lambda s.\lambda z.s\ z\\ &\bar{2} = \lambda s.\lambda z.s\ (s\ z)\\ &\bar{3} = \lambda s.\lambda z.s\ (s\ (s\ z))\\ & ...\\ &\bar{n} = \lambda s.\lambda z.s^n\ z\\ \end{align*} \]

Define z and s

• The representation \(\bar{n}\) iterates its first argument \(n\) times on its second argument.

\[ \bar{n}fx = f^b(x) \]

• To define a successor function that satisfies succ \(\bar{n} = \overline{n+1}\), we need to define \(s\) and \(z\).

\[ \begin{align*} &zeor = \bar{0} = \lambda s.\lambda z.z\\ &succ = \lambda n.\overline{n+1} = \lambda n.\lambda s.\lambda z.s(n\ s\ z)\\ \end{align*} \]

• To define addition, we need to define \(add\).

\[ \begin{align*} &add = \lambda m.\lambda n.m\ succ\ n\\ \end{align*} \]

• To define multiplication, we need to define \(mul\).

\[ \begin{align*} &mul = λm. λn. λf. m (n f)\\ \end{align*} \]

Let's walk through a concrete example of the multiplication function in lambda calculus, using Church numerals to represent numbers. The Church numeral for a natural number \( n \) is defined as a function that applies a given function \( f \) to an argument \( x \), \( n \)-times. Here’s a quick refresher on the Church numerals:

• 0: \( \lambda f. \lambda x. x \)
• 1: \( \lambda f. \lambda x. f(x) \)
• 2: \( \lambda f. \lambda x. f(f(x)) \)
• 3: \( \lambda f. \lambda x. f(f(f(x))) \)
• And so on.

Multiplication Function: The multiplication function is: [ mul = \lambda m.\lambda n.\lambda f. m (n f) ] This means \( m \) is applied \( n \)-times to a function \( f \).

Example: Multiply 2 and 3 To demonstrate, we'll compute \( 2 \times 3 \) using Church numerals for 2 and 3.

• \( 2 \) is \( \lambda f. \lambda x. f(f(x)) \)
• \( 3 \) is \( \lambda f. \lambda x. f(f(f(x))) \)

Now, let’s compute \( mul(2)(3) \):

1. Apply \( mul \) to 2:

\[ \begin{align*} &mul\ 2 = \lambda n.\lambda f. 2 (n f)\\ &\text{Substituting } 2 = \lambda f.\lambda x.f(f(x)):\\ &mul\ 2 = \lambda n.\lambda f. (\lambda f.\lambda x. f(f(x))) (n f)\\ \end{align*} \]

1. Apply \( mul(2) \) to 3:

\[ \begin{align*} &\text{Now,apply} n = 3 = \lambda f. \lambda x. f(f(f(x)))\\ &mul\ 2\ 3 = \lambda f. (\lambda x. f(f(x))) (3 f)\\ &\text{Substituting } 3 f = \lambda x. f(f(f(x))):\\ &mul\ 2\ 3 = \lambda f. (\lambda x. f(f(x))) (\lambda x. f(f(f(x))))\\ \end{align*} \]

1. Simplify the Expression:

\[ \begin{align*} &\text{Now, apply } \lambda x. f(f(x)) \text{ to } \lambda x. f(f(f(x))), \text{ replacing } x \text{ in } f(f(x)) \text{ with } \lambda x. f(f(f(x))), \text{ resulting in:}\\ &mul\ 2\ 3 = \lambda f. \lambda x. f(f(f(f(f(f(x))))))\\ \end{align*} \]

Thus, \( mul\ 2\ 3 = \lambda f. \lambda x. f(f(f(f(f(f(x)))))) \), which is the Church numeral for 6.

Conclusion: This result shows that applying the multiplication function to the Church numerals for 2 and 3 correctly gives the Church numeral for 6, as expected.

• To define exponentiation, we need to define \(exp\).

\[ \begin{align*} &exp = \lambda m.\lambda n.n\ m\\ \end{align*} \]

In lambda calculus, exponentiation (exp) can be defined as repeated application of multiplication. Specifically, exponentiation can be thought of as applying a number \( n \) (the exponent) to a function \( m \) (the base) a number of times. This means we can define exponentiation in terms of function application.

Explanation: - \( m \) is the base, which is a Church numeral. - \( n \) is the exponent, which is also a Church numeral. - In lambda calculus, \( n\ m \) means applying \( m \), \( n \)-times. Since a Church numeral represents the number of times to apply a function, this effectively results in exponentiation.

This function takes two Church numerals \( m \) and \( n \), and applies the base \( m \), \( n \)-times. The result is \( m^n \), which is the Church numeral representing the result of exponentiation.

Example: \( 2^3 \) Let’s compute \( 2^3 \) using this definition.

1. Church Numerals: - \( 2 = \lambda f. \lambda x. f(f(x)) \) - \( 3 = \lambda f. \lambda x. f(f(f(x))) \)

2. Exponentiation Function:

\[ \begin{align*} &exp = \lambda m. \lambda n. n\ m\\ \end{align*} \]

1. Apply \( exp \) to \( 2 \) and \( 3 \):

\[ \begin{align*} &exp\ 2\ 3 = 3\ 2\\ &\text{Substituting } 3 = \lambda f. \lambda x. f(f(f(x)))\\ &\text{this becomes:}\\ &exp\ 2\ 3 = (\lambda f. \lambda x. f(f(f(x))))\ 2\\ \end{align*} \]

1. Simplification:

\[ \begin{align*} &\text{Now, apply } 2 = \lambda f. \lambda x. f(f(x)) \text{ to } 3 = \lambda f. \lambda x. f(f(f(x))), \text{ resulting in:}\\ &exp\ 2\ 3 = \lambda x. 2(2(2(x)))\\ \end{align*} \]

Expanding this, each application of \( 2 \) adds another layer of function application:

\[ \begin{align*} &= \lambda x. 2(2(f(f(x)))\\ &= \lambda x. 2(f(f(f(f(x))))\\ &= \lambda x. f(f(f(f(f(f(f(f(x)))))))\\ \end{align*} \]

This result corresponds to the Church numeral for \( 8 \), which is \( 2^3 \).

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最后更新: 2024年9月19日 13:54:42
创建日期: 2024年9月19日 13:54:42