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What is the proper way to read predicate logic?



The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraIs identity included in the “key” in predicate logic?What is a formal definition of “predicate logic”?predicate logic truth valuePredicate logic, linear relationWhat is a predicate exactly in predicate logic?Predicate logic example..Predicate Logic - Simple explanation neededHow to efficiently read a predicate logic formula (best practices)Question on proving validity in predicate logicpredicate logic statements, discrete math










1












$begingroup$


Basically, in predicate logic, do we read from the inside outwards? In the example question, would 1a) be read as "For all values of y, there exists a value of x which divides y"? I've been told different things from different lecturers and online, I'm not quite sure what to do.



For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)



Any help is appreciated! Thanks



example question










share|cite|improve this question











$endgroup$











  • $begingroup$
    "There is an $y$ such that, for every $x$, ($x$ divides $y$)". In other words : is it true that there is a number (call it $y$) that is divided by every number ?
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 31 at 8:31















1












$begingroup$


Basically, in predicate logic, do we read from the inside outwards? In the example question, would 1a) be read as "For all values of y, there exists a value of x which divides y"? I've been told different things from different lecturers and online, I'm not quite sure what to do.



For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)



Any help is appreciated! Thanks



example question










share|cite|improve this question











$endgroup$











  • $begingroup$
    "There is an $y$ such that, for every $x$, ($x$ divides $y$)". In other words : is it true that there is a number (call it $y$) that is divided by every number ?
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 31 at 8:31













1












1








1





$begingroup$


Basically, in predicate logic, do we read from the inside outwards? In the example question, would 1a) be read as "For all values of y, there exists a value of x which divides y"? I've been told different things from different lecturers and online, I'm not quite sure what to do.



For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)



Any help is appreciated! Thanks



example question










share|cite|improve this question











$endgroup$




Basically, in predicate logic, do we read from the inside outwards? In the example question, would 1a) be read as "For all values of y, there exists a value of x which divides y"? I've been told different things from different lecturers and online, I'm not quite sure what to do.



For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)



Any help is appreciated! Thanks



example question







logic predicate-logic logic-translation






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 31 at 15:51









Henno Brandsma

116k349127




116k349127










asked Mar 31 at 8:20









sabsab

61




61











  • $begingroup$
    "There is an $y$ such that, for every $x$, ($x$ divides $y$)". In other words : is it true that there is a number (call it $y$) that is divided by every number ?
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 31 at 8:31
















  • $begingroup$
    "There is an $y$ such that, for every $x$, ($x$ divides $y$)". In other words : is it true that there is a number (call it $y$) that is divided by every number ?
    $endgroup$
    – Mauro ALLEGRANZA
    Mar 31 at 8:31















$begingroup$
"There is an $y$ such that, for every $x$, ($x$ divides $y$)". In other words : is it true that there is a number (call it $y$) that is divided by every number ?
$endgroup$
– Mauro ALLEGRANZA
Mar 31 at 8:31




$begingroup$
"There is an $y$ such that, for every $x$, ($x$ divides $y$)". In other words : is it true that there is a number (call it $y$) that is divided by every number ?
$endgroup$
– Mauro ALLEGRANZA
Mar 31 at 8:31










2 Answers
2






active

oldest

votes


















0












$begingroup$

Formulas in propositional and predicate logic are defined recursively. You can decompose them into terms and logical connectives by writing them as a Beth tree.



To answer your question, we read from left to right, but due to the recursive definition things can get a bit messy when conjunctions/disjunctions are involved (which is not the case in your examples). In your case 1c, we start from the left, so there is some $y$ (we fix this in the back of our mind). Now we are given any $x$ (this is the $forall x$ part). In particular, $y + 1$ could be chosen for $x$. But it's clear that $P(y+1,y)$ is false for all positive integers. Hence the sentence is false.




For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)




Yes. As an aid, think of $exists$ as "I pick a specific number" and $forall$ as "You give me any number you like". Then $exists y forall x P(x,y)$ turns into "I pick some $x$, you give me any $y$ you like, and $x$ divides $y$".






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    You read neither from inside to outside or vice versa. Rather, you have to understand how to parse the syntax first. "$∀x ( P(x) )$" means "for every $x$ it is true that $P(x)$", where $P(x)$ can be any sentence about $x$, which may include its own quantifiers. The outer brackets are there to show you what the quantifier "$∀x$" governs. If you're just starting out, you should always write the brackets. For example, think carefully what "$∀x ( ∀y ( x=y ∨ ¬∃z( x=z ∧ y=z ) ) )$" means, based on what I said about the brackets.



    It is necessary for you to first understand what the quantifier syntax I described above means, before you move on to other syntax like the one in your question. The reason is that the underlying structure is the same; you must be able to identify exactly what each quantifier governs. Consider why we can omit some brackets in the above example. "$x=z ∧ y=z$" actually means "$(x=z) ∧ (y=z)$", but why? It's because we stipulate some precedence rules, namely that we 'evaluate' operations with higher precedence first before those with lower precedence. Conventionally, the precedence rules for boolean operations and equality is:




    (highest to lowest) $=,¬,∧,∨,⇒$.




    Similarly we could include precedence rules for quantifiers, but I personally don't recommend dropping any brackets except for "$∀x ∃y ( P(x,y) )$" meaning "$∀x ( ∃y( P(x,y) ) )$". But for the sake of reading what others write, the precedence rule typically is that each quantifier governs the shortest possible part following it, so the last example becomes just "$∀x ∃y P(x,y)$".






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Formulas in propositional and predicate logic are defined recursively. You can decompose them into terms and logical connectives by writing them as a Beth tree.



      To answer your question, we read from left to right, but due to the recursive definition things can get a bit messy when conjunctions/disjunctions are involved (which is not the case in your examples). In your case 1c, we start from the left, so there is some $y$ (we fix this in the back of our mind). Now we are given any $x$ (this is the $forall x$ part). In particular, $y + 1$ could be chosen for $x$. But it's clear that $P(y+1,y)$ is false for all positive integers. Hence the sentence is false.




      For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)




      Yes. As an aid, think of $exists$ as "I pick a specific number" and $forall$ as "You give me any number you like". Then $exists y forall x P(x,y)$ turns into "I pick some $x$, you give me any $y$ you like, and $x$ divides $y$".






      share|cite|improve this answer









      $endgroup$

















        0












        $begingroup$

        Formulas in propositional and predicate logic are defined recursively. You can decompose them into terms and logical connectives by writing them as a Beth tree.



        To answer your question, we read from left to right, but due to the recursive definition things can get a bit messy when conjunctions/disjunctions are involved (which is not the case in your examples). In your case 1c, we start from the left, so there is some $y$ (we fix this in the back of our mind). Now we are given any $x$ (this is the $forall x$ part). In particular, $y + 1$ could be chosen for $x$. But it's clear that $P(y+1,y)$ is false for all positive integers. Hence the sentence is false.




        For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)




        Yes. As an aid, think of $exists$ as "I pick a specific number" and $forall$ as "You give me any number you like". Then $exists y forall x P(x,y)$ turns into "I pick some $x$, you give me any $y$ you like, and $x$ divides $y$".






        share|cite|improve this answer









        $endgroup$















          0












          0








          0





          $begingroup$

          Formulas in propositional and predicate logic are defined recursively. You can decompose them into terms and logical connectives by writing them as a Beth tree.



          To answer your question, we read from left to right, but due to the recursive definition things can get a bit messy when conjunctions/disjunctions are involved (which is not the case in your examples). In your case 1c, we start from the left, so there is some $y$ (we fix this in the back of our mind). Now we are given any $x$ (this is the $forall x$ part). In particular, $y + 1$ could be chosen for $x$. But it's clear that $P(y+1,y)$ is false for all positive integers. Hence the sentence is false.




          For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)




          Yes. As an aid, think of $exists$ as "I pick a specific number" and $forall$ as "You give me any number you like". Then $exists y forall x P(x,y)$ turns into "I pick some $x$, you give me any $y$ you like, and $x$ divides $y$".






          share|cite|improve this answer









          $endgroup$



          Formulas in propositional and predicate logic are defined recursively. You can decompose them into terms and logical connectives by writing them as a Beth tree.



          To answer your question, we read from left to right, but due to the recursive definition things can get a bit messy when conjunctions/disjunctions are involved (which is not the case in your examples). In your case 1c, we start from the left, so there is some $y$ (we fix this in the back of our mind). Now we are given any $x$ (this is the $forall x$ part). In particular, $y + 1$ could be chosen for $x$. But it's clear that $P(y+1,y)$ is false for all positive integers. Hence the sentence is false.




          For example, would 1c) be "For some values of y, every value of x divides into it" (and hence is false)




          Yes. As an aid, think of $exists$ as "I pick a specific number" and $forall$ as "You give me any number you like". Then $exists y forall x P(x,y)$ turns into "I pick some $x$, you give me any $y$ you like, and $x$ divides $y$".







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 31 at 8:43









          MacRanceMacRance

          1826




          1826





















              0












              $begingroup$

              You read neither from inside to outside or vice versa. Rather, you have to understand how to parse the syntax first. "$∀x ( P(x) )$" means "for every $x$ it is true that $P(x)$", where $P(x)$ can be any sentence about $x$, which may include its own quantifiers. The outer brackets are there to show you what the quantifier "$∀x$" governs. If you're just starting out, you should always write the brackets. For example, think carefully what "$∀x ( ∀y ( x=y ∨ ¬∃z( x=z ∧ y=z ) ) )$" means, based on what I said about the brackets.



              It is necessary for you to first understand what the quantifier syntax I described above means, before you move on to other syntax like the one in your question. The reason is that the underlying structure is the same; you must be able to identify exactly what each quantifier governs. Consider why we can omit some brackets in the above example. "$x=z ∧ y=z$" actually means "$(x=z) ∧ (y=z)$", but why? It's because we stipulate some precedence rules, namely that we 'evaluate' operations with higher precedence first before those with lower precedence. Conventionally, the precedence rules for boolean operations and equality is:




              (highest to lowest) $=,¬,∧,∨,⇒$.




              Similarly we could include precedence rules for quantifiers, but I personally don't recommend dropping any brackets except for "$∀x ∃y ( P(x,y) )$" meaning "$∀x ( ∃y( P(x,y) ) )$". But for the sake of reading what others write, the precedence rule typically is that each quantifier governs the shortest possible part following it, so the last example becomes just "$∀x ∃y P(x,y)$".






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                You read neither from inside to outside or vice versa. Rather, you have to understand how to parse the syntax first. "$∀x ( P(x) )$" means "for every $x$ it is true that $P(x)$", where $P(x)$ can be any sentence about $x$, which may include its own quantifiers. The outer brackets are there to show you what the quantifier "$∀x$" governs. If you're just starting out, you should always write the brackets. For example, think carefully what "$∀x ( ∀y ( x=y ∨ ¬∃z( x=z ∧ y=z ) ) )$" means, based on what I said about the brackets.



                It is necessary for you to first understand what the quantifier syntax I described above means, before you move on to other syntax like the one in your question. The reason is that the underlying structure is the same; you must be able to identify exactly what each quantifier governs. Consider why we can omit some brackets in the above example. "$x=z ∧ y=z$" actually means "$(x=z) ∧ (y=z)$", but why? It's because we stipulate some precedence rules, namely that we 'evaluate' operations with higher precedence first before those with lower precedence. Conventionally, the precedence rules for boolean operations and equality is:




                (highest to lowest) $=,¬,∧,∨,⇒$.




                Similarly we could include precedence rules for quantifiers, but I personally don't recommend dropping any brackets except for "$∀x ∃y ( P(x,y) )$" meaning "$∀x ( ∃y( P(x,y) ) )$". But for the sake of reading what others write, the precedence rule typically is that each quantifier governs the shortest possible part following it, so the last example becomes just "$∀x ∃y P(x,y)$".






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  You read neither from inside to outside or vice versa. Rather, you have to understand how to parse the syntax first. "$∀x ( P(x) )$" means "for every $x$ it is true that $P(x)$", where $P(x)$ can be any sentence about $x$, which may include its own quantifiers. The outer brackets are there to show you what the quantifier "$∀x$" governs. If you're just starting out, you should always write the brackets. For example, think carefully what "$∀x ( ∀y ( x=y ∨ ¬∃z( x=z ∧ y=z ) ) )$" means, based on what I said about the brackets.



                  It is necessary for you to first understand what the quantifier syntax I described above means, before you move on to other syntax like the one in your question. The reason is that the underlying structure is the same; you must be able to identify exactly what each quantifier governs. Consider why we can omit some brackets in the above example. "$x=z ∧ y=z$" actually means "$(x=z) ∧ (y=z)$", but why? It's because we stipulate some precedence rules, namely that we 'evaluate' operations with higher precedence first before those with lower precedence. Conventionally, the precedence rules for boolean operations and equality is:




                  (highest to lowest) $=,¬,∧,∨,⇒$.




                  Similarly we could include precedence rules for quantifiers, but I personally don't recommend dropping any brackets except for "$∀x ∃y ( P(x,y) )$" meaning "$∀x ( ∃y( P(x,y) ) )$". But for the sake of reading what others write, the precedence rule typically is that each quantifier governs the shortest possible part following it, so the last example becomes just "$∀x ∃y P(x,y)$".






                  share|cite|improve this answer









                  $endgroup$



                  You read neither from inside to outside or vice versa. Rather, you have to understand how to parse the syntax first. "$∀x ( P(x) )$" means "for every $x$ it is true that $P(x)$", where $P(x)$ can be any sentence about $x$, which may include its own quantifiers. The outer brackets are there to show you what the quantifier "$∀x$" governs. If you're just starting out, you should always write the brackets. For example, think carefully what "$∀x ( ∀y ( x=y ∨ ¬∃z( x=z ∧ y=z ) ) )$" means, based on what I said about the brackets.



                  It is necessary for you to first understand what the quantifier syntax I described above means, before you move on to other syntax like the one in your question. The reason is that the underlying structure is the same; you must be able to identify exactly what each quantifier governs. Consider why we can omit some brackets in the above example. "$x=z ∧ y=z$" actually means "$(x=z) ∧ (y=z)$", but why? It's because we stipulate some precedence rules, namely that we 'evaluate' operations with higher precedence first before those with lower precedence. Conventionally, the precedence rules for boolean operations and equality is:




                  (highest to lowest) $=,¬,∧,∨,⇒$.




                  Similarly we could include precedence rules for quantifiers, but I personally don't recommend dropping any brackets except for "$∀x ∃y ( P(x,y) )$" meaning "$∀x ( ∃y( P(x,y) ) )$". But for the sake of reading what others write, the precedence rule typically is that each quantifier governs the shortest possible part following it, so the last example becomes just "$∀x ∃y P(x,y)$".







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 31 at 10:33









                  user21820user21820

                  40.1k544162




                  40.1k544162



























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