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Why do ¬, ∀ and ∃ have the same precedence?
Why do the sequent calculus NOT left and NOT right rules work?Why ⊢ for affirmative predicates and ⊨ for ¬negations?Difference between First Order Logic and Predicate CalculusWhat is the difference between superposition and paramodulation?Precedence of satisfiability operatorWhy does soundness imply consistency?Can proof by contradiction work without the law of excluded middle?Logical and non logical symbols and predicatesTrying to understand interpretation and denotation in FOLWhy the given statement can't be expressed using predicates and quantifiers in the way described in details?
$begingroup$
I thought the order of precedence of operators and quantifiers was arbitrary, but I don't really understand why those three have the same "strength" in relation to other operators (e.g., ¬ will have precedence over ∧, but not over ∀). This leads to the rule being that ¬, ∀ and ∃ will bind to the closest predicate on their right (if I understood correctly). Why is this?
first-order-logic
asked 11 hours ago
PhilPhil
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I thought the order of precedence of operators and quantifiers was arbitrary, but I don't really understand why those three have the same "strength" in relation to other operators (e.g., ¬ will have precedence over ∧, but not over ∀). This leads to the rule being that ¬, ∀ and ∃ will bind to the closest predicate on their right (if I understood correctly). Why is this?
first-order-logic
asked 11 hours ago
PhilPhil
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
16
$begingroup$
I would be very wary of ideas of precedence in first-order logic. Don't think you can write $forall x,Alor B$ or $Alor Bland C$ and have people understand what you mean.
$endgroup$
– David Richerby
10 hours ago
3
$begingroup$
As far as I can see, precedence of logical connectives and quantifiers can vary from author to author. I tend to write $forall x. (p(x) implies q(x))$ without parentheses, like several others. Some instead write $(forall x. p(x)) implies q$ without parentheses, using the opposite precedence. I learned to never take precedence for granted, and sometimes to use a few redundant parentheses to be sure to be understood by everyone.
$endgroup$
– chi
6 hours ago
add a comment |
$begingroup$
I thought the order of precedence of operators and quantifiers was arbitrary, but I don't really understand why those three have the same "strength" in relation to other operators (e.g., ¬ will have precedence over ∧, but not over ∀). This leads to the rule being that ¬, ∀ and ∃ will bind to the closest predicate on their right (if I understood correctly). Why is this?
first-order-logic
asked 11 hours ago
PhilPhil
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I thought the order of precedence of operators and quantifiers was arbitrary, but I don't really understand why those three have the same "strength" in relation to other operators (e.g., ¬ will have precedence over ∧, but not over ∀). This leads to the rule being that ¬, ∀ and ∃ will bind to the closest predicate on their right (if I understood correctly). Why is this?
first-order-logic
first-order-logic
asked 11 hours ago
PhilPhil
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
PhilPhil
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
Phil
asked 11 hours ago
PhilPhil
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
PhilPhil
212
asked 11 hours ago
PhilPhil
212
212
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Phil is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
16
$begingroup$
I would be very wary of ideas of precedence in first-order logic. Don't think you can write $forall x,Alor B$ or $Alor Bland C$ and have people understand what you mean.
$endgroup$
– David Richerby
10 hours ago
3
$begingroup$
As far as I can see, precedence of logical connectives and quantifiers can vary from author to author. I tend to write $forall x. (p(x) implies q(x))$ without parentheses, like several others. Some instead write $(forall x. p(x)) implies q$ without parentheses, using the opposite precedence. I learned to never take precedence for granted, and sometimes to use a few redundant parentheses to be sure to be understood by everyone.
$endgroup$
– chi
6 hours ago
add a comment |
16
$begingroup$
I would be very wary of ideas of precedence in first-order logic. Don't think you can write $forall x,Alor B$ or $Alor Bland C$ and have people understand what you mean.
$endgroup$
– David Richerby
10 hours ago
3
$begingroup$
As far as I can see, precedence of logical connectives and quantifiers can vary from author to author. I tend to write $forall x. (p(x) implies q(x))$ without parentheses, like several others. Some instead write $(forall x. p(x)) implies q$ without parentheses, using the opposite precedence. I learned to never take precedence for granted, and sometimes to use a few redundant parentheses to be sure to be understood by everyone.
$endgroup$
– chi
6 hours ago
16
16
$begingroup$
I would be very wary of ideas of precedence in first-order logic. Don't think you can write $forall x,Alor B$ or $Alor Bland C$ and have people understand what you mean.
$endgroup$
– David Richerby
10 hours ago
$begingroup$
I would be very wary of ideas of precedence in first-order logic. Don't think you can write $forall x,Alor B$ or $Alor Bland C$ and have people understand what you mean.
$endgroup$
– David Richerby
10 hours ago
3
3
$begingroup$
As far as I can see, precedence of logical connectives and quantifiers can vary from author to author. I tend to write $forall x. (p(x) implies q(x))$ without parentheses, like several others. Some instead write $(forall x. p(x)) implies q$ without parentheses, using the opposite precedence. I learned to never take precedence for granted, and sometimes to use a few redundant parentheses to be sure to be understood by everyone.
$endgroup$
– chi
6 hours ago
$begingroup$
As far as I can see, precedence of logical connectives and quantifiers can vary from author to author. I tend to write $forall x. (p(x) implies q(x))$ without parentheses, like several others. Some instead write $(forall x. p(x)) implies q$ without parentheses, using the opposite precedence. I learned to never take precedence for granted, and sometimes to use a few redundant parentheses to be sure to be understood by everyone.
$endgroup$
– chi
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Order of precedence is simply a notional convenience. There is no notion of strength here, just notation. All three operators are unary operators with notation "$circ cdot$", where $circ$ denotes the operator symbol $exists, forall,neg$ and $cdot$ the operand. There can never be any ambiguity in which order to apply these operators: the operator to the right must always be applied to the operand first.
Hence, they have the same precedence among eachother if we consider only those three operators. (Note that there can be ambiguity if the unary operators have different position, e.g. $-x^2$, this could mean either $(-x)^2$ or $-(x^2)$ if there was no precendence between $^2$ and $-$.)
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
$endgroup$
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
add a comment |
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$begingroup$
Order of precedence is simply a notional convenience. There is no notion of strength here, just notation. All three operators are unary operators with notation "$circ cdot$", where $circ$ denotes the operator symbol $exists, forall,neg$ and $cdot$ the operand. There can never be any ambiguity in which order to apply these operators: the operator to the right must always be applied to the operand first.
Hence, they have the same precedence among eachother if we consider only those three operators. (Note that there can be ambiguity if the unary operators have different position, e.g. $-x^2$, this could mean either $(-x)^2$ or $-(x^2)$ if there was no precendence between $^2$ and $-$.)
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
$endgroup$
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
add a comment |
$begingroup$
Order of precedence is simply a notional convenience. There is no notion of strength here, just notation. All three operators are unary operators with notation "$circ cdot$", where $circ$ denotes the operator symbol $exists, forall,neg$ and $cdot$ the operand. There can never be any ambiguity in which order to apply these operators: the operator to the right must always be applied to the operand first.
Hence, they have the same precedence among eachother if we consider only those three operators. (Note that there can be ambiguity if the unary operators have different position, e.g. $-x^2$, this could mean either $(-x)^2$ or $-(x^2)$ if there was no precendence between $^2$ and $-$.)
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
$endgroup$
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
add a comment |
$begingroup$
Order of precedence is simply a notional convenience. There is no notion of strength here, just notation. All three operators are unary operators with notation "$circ cdot$", where $circ$ denotes the operator symbol $exists, forall,neg$ and $cdot$ the operand. There can never be any ambiguity in which order to apply these operators: the operator to the right must always be applied to the operand first.
Hence, they have the same precedence among eachother if we consider only those three operators. (Note that there can be ambiguity if the unary operators have different position, e.g. $-x^2$, this could mean either $(-x)^2$ or $-(x^2)$ if there was no precendence between $^2$ and $-$.)
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
$endgroup$
Order of precedence is simply a notional convenience. There is no notion of strength here, just notation. All three operators are unary operators with notation "$circ cdot$", where $circ$ denotes the operator symbol $exists, forall,neg$ and $cdot$ the operand. There can never be any ambiguity in which order to apply these operators: the operator to the right must always be applied to the operand first.
Hence, they have the same precedence among eachother if we consider only those three operators. (Note that there can be ambiguity if the unary operators have different position, e.g. $-x^2$, this could mean either $(-x)^2$ or $-(x^2)$ if there was no precendence between $^2$ and $-$.)
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
edited 8 hours ago
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
answered 11 hours ago
Discrete lizard♦Discrete lizard
4,39411537
4,39411537
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
add a comment |
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
$begingroup$
The real question is why you would want the quantifiers to bind so tightly, assuming that most statements will be in prenex normal form and therefore need parentheses.
$endgroup$
– Kevin
3 hours ago
add a comment |
Phil is a new contributor. Be nice, and check out our Code of Conduct.
Phil is a new contributor. Be nice, and check out our Code of Conduct.
Phil is a new contributor. Be nice, and check out our Code of Conduct.
Phil is a new contributor. Be nice, and check out our Code of Conduct.
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16
$begingroup$
I would be very wary of ideas of precedence in first-order logic. Don't think you can write $forall x,Alor B$ or $Alor Bland C$ and have people understand what you mean.
$endgroup$
– David Richerby
10 hours ago
3
$begingroup$
As far as I can see, precedence of logical connectives and quantifiers can vary from author to author. I tend to write $forall x. (p(x) implies q(x))$ without parentheses, like several others. Some instead write $(forall x. p(x)) implies q$ without parentheses, using the opposite precedence. I learned to never take precedence for granted, and sometimes to use a few redundant parentheses to be sure to be understood by everyone.
$endgroup$
– chi
6 hours ago