3 Copy-edit and formatting

I have started learning lambda calculus from the book by Hindley and Seldin  .

It brought up the concept with the function 'x-y'First$$x-y$$, first as a function of "x"$$x$$ and then as a function of "y"$$y$$.In In a way  , it emphasized the difference between treating the expression as a function of "x:$$x$$ and function of "y"$$y$$,giving giving separate names to the functions  .

$$f(x,y) = x -y$$ and $$g(y,x) = x -y$$$$f(x,y) = x-y \qquad\text{and}\qquad g(y,x) = x -y$$ and, in $$\lambda$$ notion

$$h = \lambda x.x-y$$ and $$g = \lambda y x. x-y$$, $$h = \lambda xy.x-y \qquad\text{and}\qquad g = \lambda yx. x-y\,.$$

If I am not wrong then it can be inferred that :

$$f =\lambda x .(x-y)$$$$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with "x" $$x$$ and similarly when "lambda "$$\lambda$$ is placed with $$x$$$$x$$ the other variable $$y$$" will be kept as a constant  .

But then it said that this can be denoted as  ,with with $$h$$ being the common name of the function

$$h = \lambda xy.(x-y)$$ and $$h = \lambda yx.(x-y)$$ $$h = \lambda xy.(x-y)\qquad\text{and}\qquad h = \lambda yx.(x-y)\,.$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant  ? The above function is called a two-place function ,whatfunction; what does itthat mean  ? Then

Then he introduces the following function in $$\lambda$$ notionnotation calling it a one point-point function  : $$h^\star = \lambda x .(\lambda y . x−y)$$$$h^\star = \lambda x .(\lambda y . x−y)$$ and mentions the following

Forsays that, for each number a$$a$$, we have $$h^\star(a)=λy . a−y$$ $$h^\star(a)=\lambda y . a−y$$. Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$  . Why sois this? And

And then it deduces the followingthat :

$$(h^\star(a))(b)=(\lambda y . a−y)(b) = a−b = h(a,b)$$ and says $$h^\star$$ can be viewed as 'representing '"representing" $$h$$  . Is "representing" a technical term  ?How How was $$(\lambda y . a−y)(b) = a−b$$ concluded .

P.S : I am afraid this question may be classified as unclear or broad but I hope that this can be unified as a problem of getting though the $$\lambda$$ notion or missing something and the few questions asked at different parts if this post will have very closely related answer.?

I have started learning lambda calculus from the book by Hindley and Seldin  .

It brought up the concept with the function 'x-y'First as a function of "x" and then as a function of "y".In a way  , it emphasized the difference between treating the expression as a function of "x: and function of "y",giving separate names to the functions  .

$$f(x,y) = x -y$$ and $$g(y,x) = x -y$$ and in $$\lambda$$ notion

$$h = \lambda x.x-y$$ and $$g = \lambda y x. x-y$$

If I am not wrong then it can be inferred that :

$$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with "x" and similarly when "lambda " is placed with $$x$$ the other variable $$y$$" will be kept as a constant  .

But then it said that this can be denoted as  ,with $$h$$ being the common name of the function

$$h = \lambda xy.(x-y)$$ and $$h = \lambda yx.(x-y)$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant  ? The above function is called a two-place function ,what does it mean  ? Then he introduces the following function in $$\lambda$$ notion calling it a one point function  : $$h^\star = \lambda x .(\lambda y . x−y)$$ and mentions the following

For each number a, we have $$h^\star(a)=λy . a−y$$ . Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$  . Why so? And then it deduces the following :

$$(h^\star(a))(b)=(\lambda y . a−y)(b) = a−b = h(a,b)$$ and says $$h^\star$$ can be viewed as 'representing ' $$h$$  . Is "representing" a technical term  ?How was $$(\lambda y . a−y)(b) = a−b$$ concluded .

P.S : I am afraid this question may be classified as unclear or broad but I hope that this can be unified as a problem of getting though the $$\lambda$$ notion or missing something and the few questions asked at different parts if this post will have very closely related answer.

I have started learning lambda calculus from the book by Hindley and Seldin.

It brought up the concept with the function $$x-y$$, first as a function of $$x$$ and then as a function of $$y$$. In a way, it emphasized the difference between treating the expression as a function of $$x$$ and function of $$y$$, giving separate names to the functions.

$$f(x,y) = x-y \qquad\text{and}\qquad g(y,x) = x -y$$ and, in $$\lambda$$ notion, $$h = \lambda xy.x-y \qquad\text{and}\qquad g = \lambda yx. x-y\,.$$

If I am not wrong then it can be inferred that $$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with $$x$$ and similarly when $$\lambda$$ is placed with $$x$$ the other variable $$y$$ will be kept as a constant.

But then it said that this can be denoted as, with $$h$$ being the common name of the function $$h = \lambda xy.(x-y)\qquad\text{and}\qquad h = \lambda yx.(x-y)\,.$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant? The above function is called a two-place function; what does that mean?

Then he introduces the following function in $$\lambda$$ notation calling it a one-point function: $$h^\star = \lambda x .(\lambda y . x−y)$$ and says that, for each number $$a$$, we have $$h^\star(a)=\lambda y . a−y$$. Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$. Why is this?

And then it deduces that $$(h^\star(a))(b)=(\lambda y . a−y)(b) = a−b = h(a,b)$$ and says $$h^\star$$ can be viewed as "representing" $$h$$. Is "representing" a technical term? How was $$(\lambda y . a−y)(b) = a−b$$ concluded?

2 Fixed h "exponent" in LaTeX

I have started learning lambda calculus from the book by Hindley and Seldin .

It brought up the concept with the function 'x-y'First as a function of "x" and then as a function of "y".In a way , it emphasized the difference between treating the expression as a function of "x: and function of "y",giving separate names to the functions .

$$f(x,y) = x -y$$ and $$g(y,x) = x -y$$ and in $$\lambda$$ notion

$$h = \lambda x.x-y$$ and $$g = \lambda yx. x-y$$$$g = \lambda y x. x-y$$

If I am not wrong then it can be inferred that :

$$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with "x" and similarly when "lambda " is placed with $$x$$ the other variable $$y$$" will be kept as a constant .

But then it said that this can be denoted as ,with $$h$$ being the common name of the function

$$h = \lambda xy.(x-y)$$ and $$h = \lambda yx.(x-y)$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant ? The above function is called a two-place function ,what does it mean ? Then he introduces the following function in $$\lambda$$ notion calling it a one point function : $$h* = λx .(λy . x−y)$$$$h^\star = \lambda x .(\lambda y . x−y)$$ and mentions the following

For each number a, we have $$h*(a)=λy . a−y$$$$h^\star(a)=λy . a−y$$ . Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$ . Why so  ? And then it deduces the following : $$(h*(a))(b)=(λy . a−y)(b) = a−b = h(a,b)$$

$$(h^\star(a))(b)=(\lambda y . a−y)(b) = a−b = h(a,b)$$ and says $$h*$$$$h^\star$$ can be viewed as 'representing ' $$h$$ . Is "representing" a technical term ?How was $$(λy . a−y)(b) = a−b$$$$(\lambda y . a−y)(b) = a−b$$ concluded .

P.S : I am afraid this question may be classified as unclear or broad but I hope that this can be unified as a problem of getting though the $$\lambda$$ notion or missing something and the few questions asked at different parts if this post will have very closely related answer.

I have started learning lambda calculus from the book by Hindley and Seldin .

It brought up the concept with the function 'x-y'First as a function of "x" and then as a function of "y".In a way , it emphasized the difference between treating the expression as a function of "x: and function of "y",giving separate names to the functions .

$$f(x,y) = x -y$$ and $$g(y,x) = x -y$$ and in $$\lambda$$ notion

$$h = \lambda x.x-y$$ and $$g = \lambda yx. x-y$$

If I am not wrong then it can be inferred that :

$$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with "x" and similarly when "lambda " is placed with $$x$$ the other variable $$y$$" will be kept as a constant .

But then it said that this can be denoted as ,with $$h$$ being the common name of the function

$$h = \lambda xy.(x-y)$$ and $$h = \lambda yx.(x-y)$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant ? The above function is called a two-place function ,what does it mean ? Then he introduces the following function in $$\lambda$$ notion calling it a one point function : $$h* = λx .(λy . x−y)$$ and mentions the following

For each number a, we have $$h*(a)=λy . a−y$$ . Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$ . Why so  ? And then it deduces the following : $$(h*(a))(b)=(λy . a−y)(b) = a−b = h(a,b)$$ and says $$h*$$ can be viewed as 'representing ' $$h$$ . Is "representing" a technical term ?How was $$(λy . a−y)(b) = a−b$$ concluded .

P.S : I am afraid this question may be classified as unclear or broad but I hope that this can be unified as a problem of getting though the $$\lambda$$ notion or missing something and the few questions asked at different parts if this post will have very closely related answer.

I have started learning lambda calculus from the book by Hindley and Seldin .

It brought up the concept with the function 'x-y'First as a function of "x" and then as a function of "y".In a way , it emphasized the difference between treating the expression as a function of "x: and function of "y",giving separate names to the functions .

$$f(x,y) = x -y$$ and $$g(y,x) = x -y$$ and in $$\lambda$$ notion

$$h = \lambda x.x-y$$ and $$g = \lambda y x. x-y$$

If I am not wrong then it can be inferred that :

$$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with "x" and similarly when "lambda " is placed with $$x$$ the other variable $$y$$" will be kept as a constant .

But then it said that this can be denoted as ,with $$h$$ being the common name of the function

$$h = \lambda xy.(x-y)$$ and $$h = \lambda yx.(x-y)$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant ? The above function is called a two-place function ,what does it mean ? Then he introduces the following function in $$\lambda$$ notion calling it a one point function : $$h^\star = \lambda x .(\lambda y . x−y)$$ and mentions the following

For each number a, we have $$h^\star(a)=λy . a−y$$ . Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$ . Why so? And then it deduces the following :

$$(h^\star(a))(b)=(\lambda y . a−y)(b) = a−b = h(a,b)$$ and says $$h^\star$$ can be viewed as 'representing ' $$h$$ . Is "representing" a technical term ?How was $$(\lambda y . a−y)(b) = a−b$$ concluded .

P.S : I am afraid this question may be classified as unclear or broad but I hope that this can be unified as a problem of getting though the $$\lambda$$ notion or missing something and the few questions asked at different parts if this post will have very closely related answer.

1

# Basic notation in lambda calculus

I have started learning lambda calculus from the book by Hindley and Seldin .

It brought up the concept with the function 'x-y'First as a function of "x" and then as a function of "y".In a way , it emphasized the difference between treating the expression as a function of "x: and function of "y",giving separate names to the functions .

$$f(x,y) = x -y$$ and $$g(y,x) = x -y$$ and in $$\lambda$$ notion

$$h = \lambda x.x-y$$ and $$g = \lambda yx. x-y$$

If I am not wrong then it can be inferred that :

$$f =\lambda x .(x-y)$$ means that while computing the value of the expression the expression will vary only with "x" and similarly when "lambda " is placed with $$x$$ the other variable $$y$$" will be kept as a constant .

But then it said that this can be denoted as ,with $$h$$ being the common name of the function

$$h = \lambda xy.(x-y)$$ and $$h = \lambda yx.(x-y)$$
Is it so that in the above expression the variable which is adjacent to $$\lambda$$ receives the value and the other one is constant ? The above function is called a two-place function ,what does it mean ? Then he introduces the following function in $$\lambda$$ notion calling it a one point function : $$h* = λx .(λy . x−y)$$ and mentions the following

For each number a, we have $$h*(a)=λy . a−y$$ . Here $$a$$ is being provided as an argument but $$\lambda$$ is placed near $$y$$ . Why so ? And then it deduces the following : $$(h*(a))(b)=(λy . a−y)(b) = a−b = h(a,b)$$ and says $$h*$$ can be viewed as 'representing ' $$h$$ . Is "representing" a technical term ?How was $$(λy . a−y)(b) = a−b$$ concluded .

P.S : I am afraid this question may be classified as unclear or broad but I hope that this can be unified as a problem of getting though the $$\lambda$$ notion or missing something and the few questions asked at different parts if this post will have very closely related answer.