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lambda calculus precedence of application and abstraction
How helpful is knowing lambda calculus?What are some resources for learning Lambda Calculus?Convert Python to Haskell / Lambda calculusWhat type of lambda calculus would Lisp loosely be an example of?Lambda Calculus operators precedenceCode exercising the unique possibilities of each edge of the lambda calculusWhich FP language follows lambda calculus the closest?“What part of Hindley-Milner do you not understand?”Quantifiers in lambda calculusEta abstraction in lambda calculus
Application has higher precedence than abstraction.
In this sense, what is lambda calculus abstraction? I'm confused at what there is to have precedence over?
apply abstraction operator-precedence lambda-calculus
asked Mar 21 at 14:27
HdotHdot
43
add a comment |
Application has higher precedence than abstraction.
In this sense, what is lambda calculus abstraction? I'm confused at what there is to have precedence over?
apply abstraction operator-precedence lambda-calculus
asked Mar 21 at 14:27
HdotHdot
43
add a comment |
Application has higher precedence than abstraction.
In this sense, what is lambda calculus abstraction? I'm confused at what there is to have precedence over?
apply abstraction operator-precedence lambda-calculus
asked Mar 21 at 14:27
HdotHdot
43
Application has higher precedence than abstraction.
In this sense, what is lambda calculus abstraction? I'm confused at what there is to have precedence over?
apply abstraction operator-precedence lambda-calculus
apply abstraction operator-precedence lambda-calculus
asked Mar 21 at 14:27
HdotHdot
43
asked Mar 21 at 14:27
HdotHdot
43
asked Mar 21 at 14:27
HdotHdot
43
asked Mar 21 at 14:27
HdotHdot
43
asked Mar 21 at 14:27
HdotHdot
43
43
add a comment |
add a comment |
1 Answer
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Lambda abstraction is λx.M, for some variable x and arbitrary term M.
Application is (MN), for some arbitrary terms M and N.
Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic, multiplication by convention has precedence over addition, which means that 5+2×3 is read as 5+(2×3) and not as (5+2)×3. The multiplication operator is evaluated first and binds the terms closest to it, and addition comes secondary, embedding the multiplication term.
W.r.t. to lambda calculus, the convention that application has higher precedence than abstraction means that in case of doubt because brackets have been ommitted, you will first try to form an application and only afterwards perform abstraction, so application "binds" stronger, and an abstraction term will be formed later and subsume the application term.
E.g., λx.M N could in principle be read as either λx.(MN) or (λx.M)M, but since application has precedence over application, you first form the possible application (MN) and then the abstraction λx.(MN). If it were the other way round, i.e. if abstraction had precedence over application, then you would first try to form an abstraction term (λx.M), then application with the term you already got ((λx.M)M).
So by defining that application has precedence over abstraction, λx.M N = λx.(MN), and not ((λx.M)M).
answered Mar 21 at 19:51
lemontreelemontree
1177
add a comment |
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Lambda abstraction is λx.M, for some variable x and arbitrary term M.
Application is (MN), for some arbitrary terms M and N.
Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic, multiplication by convention has precedence over addition, which means that 5+2×3 is read as 5+(2×3) and not as (5+2)×3. The multiplication operator is evaluated first and binds the terms closest to it, and addition comes secondary, embedding the multiplication term.
W.r.t. to lambda calculus, the convention that application has higher precedence than abstraction means that in case of doubt because brackets have been ommitted, you will first try to form an application and only afterwards perform abstraction, so application "binds" stronger, and an abstraction term will be formed later and subsume the application term.
E.g., λx.M N could in principle be read as either λx.(MN) or (λx.M)M, but since application has precedence over application, you first form the possible application (MN) and then the abstraction λx.(MN). If it were the other way round, i.e. if abstraction had precedence over application, then you would first try to form an abstraction term (λx.M), then application with the term you already got ((λx.M)M).
So by defining that application has precedence over abstraction, λx.M N = λx.(MN), and not ((λx.M)M).
answered Mar 21 at 19:51
lemontreelemontree
1177
add a comment |
Lambda abstraction is λx.M, for some variable x and arbitrary term M.
Application is (MN), for some arbitrary terms M and N.
Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic, multiplication by convention has precedence over addition, which means that 5+2×3 is read as 5+(2×3) and not as (5+2)×3. The multiplication operator is evaluated first and binds the terms closest to it, and addition comes secondary, embedding the multiplication term.
W.r.t. to lambda calculus, the convention that application has higher precedence than abstraction means that in case of doubt because brackets have been ommitted, you will first try to form an application and only afterwards perform abstraction, so application "binds" stronger, and an abstraction term will be formed later and subsume the application term.
E.g., λx.M N could in principle be read as either λx.(MN) or (λx.M)M, but since application has precedence over application, you first form the possible application (MN) and then the abstraction λx.(MN). If it were the other way round, i.e. if abstraction had precedence over application, then you would first try to form an abstraction term (λx.M), then application with the term you already got ((λx.M)M).
So by defining that application has precedence over abstraction, λx.M N = λx.(MN), and not ((λx.M)M).
answered Mar 21 at 19:51
lemontreelemontree
1177
add a comment |
Lambda abstraction is λx.M, for some variable x and arbitrary term M.
Application is (MN), for some arbitrary terms M and N.
Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic, multiplication by convention has precedence over addition, which means that 5+2×3 is read as 5+(2×3) and not as (5+2)×3. The multiplication operator is evaluated first and binds the terms closest to it, and addition comes secondary, embedding the multiplication term.
W.r.t. to lambda calculus, the convention that application has higher precedence than abstraction means that in case of doubt because brackets have been ommitted, you will first try to form an application and only afterwards perform abstraction, so application "binds" stronger, and an abstraction term will be formed later and subsume the application term.
E.g., λx.M N could in principle be read as either λx.(MN) or (λx.M)M, but since application has precedence over application, you first form the possible application (MN) and then the abstraction λx.(MN). If it were the other way round, i.e. if abstraction had precedence over application, then you would first try to form an abstraction term (λx.M), then application with the term you already got ((λx.M)M).
So by defining that application has precedence over abstraction, λx.M N = λx.(MN), and not ((λx.M)M).
answered Mar 21 at 19:51
lemontreelemontree
1177
Lambda abstraction is λx.M, for some variable x and arbitrary term M.
Application is (MN), for some arbitrary terms M and N.
Precdence is the question which of several operation is to be performed first if more than one reading is possible because the term is ambiguous due to ommission of brackets. For example in arithmetic, multiplication by convention has precedence over addition, which means that 5+2×3 is read as 5+(2×3) and not as (5+2)×3. The multiplication operator is evaluated first and binds the terms closest to it, and addition comes secondary, embedding the multiplication term.
W.r.t. to lambda calculus, the convention that application has higher precedence than abstraction means that in case of doubt because brackets have been ommitted, you will first try to form an application and only afterwards perform abstraction, so application "binds" stronger, and an abstraction term will be formed later and subsume the application term.
E.g., λx.M N could in principle be read as either λx.(MN) or (λx.M)M, but since application has precedence over application, you first form the possible application (MN) and then the abstraction λx.(MN). If it were the other way round, i.e. if abstraction had precedence over application, then you would first try to form an abstraction term (λx.M), then application with the term you already got ((λx.M)M).
So by defining that application has precedence over abstraction, λx.M N = λx.(MN), and not ((λx.M)M).
answered Mar 21 at 19:51
lemontreelemontree
1177
answered Mar 21 at 19:51
lemontreelemontree
1177
answered Mar 21 at 19:51
lemontreelemontree
1177
answered Mar 21 at 19:51
lemontreelemontree
1177
1177
add a comment |
add a comment |
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