Rules of distribution of quantifiers over conditional and biconditional












3














Which of the following propositional logic statements are true and why?




  1. $(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))$

  2. $(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))$

  3. $(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))$

  4. $(∀x(P(x)⇔Q(x)))⟹(∀x(P(x))⇔(∀x(Q(x))))$


  5. Are their any standard laws/rules of distribution of universal quantifier over conditional and binconditional that can help me solve this?


  6. Also rules for distribution of existential quantifier over conditional and binconditional?



Recently I came across distribution of quantifiers over $vee$ and $wedge$, which gave set theoretic interpretation of them as follows:





  • $((∀x)G(x)∨ (∀x)H(x))→ (∀x)(F(x)∨ G(x))$



    In set theoretic terms,
    if we have that $(f(G) = D ∨ f(H) = D)$, then we have $(f(G) ∪ f(H)) = D$




  • $(∃x)(G(x)∧ H(x))→((∃x)G(x)∧ (∃x)H(x))$



    In set theoretic terms,
    if we have that $(f(G) ∩ f(H)) ≥ 1$, then we have $(f(G) ≥ 1 ∧ f(H) ≥ 1)$




Can we say similar for distribution of quantifiers over conditional and biconditional (just to bring in more clarity)?










share|cite|improve this question
























  • The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic.
    – Taroccoesbrocco
    Jan 1 '16 at 21:23












  • Do you want me to change the title to something else?
    – anir123
    Jan 1 '16 at 21:25












  • Yes, the title is misleading, as well as the tag "propositional calculus".
    – Taroccoesbrocco
    Jan 1 '16 at 21:27






  • 1




    Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?"
    – anir123
    Jan 1 '16 at 21:31










  • Now the title is ok for me.
    – Taroccoesbrocco
    Jan 1 '16 at 21:33
















3














Which of the following propositional logic statements are true and why?




  1. $(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))$

  2. $(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))$

  3. $(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))$

  4. $(∀x(P(x)⇔Q(x)))⟹(∀x(P(x))⇔(∀x(Q(x))))$


  5. Are their any standard laws/rules of distribution of universal quantifier over conditional and binconditional that can help me solve this?


  6. Also rules for distribution of existential quantifier over conditional and binconditional?



Recently I came across distribution of quantifiers over $vee$ and $wedge$, which gave set theoretic interpretation of them as follows:





  • $((∀x)G(x)∨ (∀x)H(x))→ (∀x)(F(x)∨ G(x))$



    In set theoretic terms,
    if we have that $(f(G) = D ∨ f(H) = D)$, then we have $(f(G) ∪ f(H)) = D$




  • $(∃x)(G(x)∧ H(x))→((∃x)G(x)∧ (∃x)H(x))$



    In set theoretic terms,
    if we have that $(f(G) ∩ f(H)) ≥ 1$, then we have $(f(G) ≥ 1 ∧ f(H) ≥ 1)$




Can we say similar for distribution of quantifiers over conditional and biconditional (just to bring in more clarity)?










share|cite|improve this question
























  • The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic.
    – Taroccoesbrocco
    Jan 1 '16 at 21:23












  • Do you want me to change the title to something else?
    – anir123
    Jan 1 '16 at 21:25












  • Yes, the title is misleading, as well as the tag "propositional calculus".
    – Taroccoesbrocco
    Jan 1 '16 at 21:27






  • 1




    Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?"
    – anir123
    Jan 1 '16 at 21:31










  • Now the title is ok for me.
    – Taroccoesbrocco
    Jan 1 '16 at 21:33














3












3








3


1





Which of the following propositional logic statements are true and why?




  1. $(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))$

  2. $(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))$

  3. $(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))$

  4. $(∀x(P(x)⇔Q(x)))⟹(∀x(P(x))⇔(∀x(Q(x))))$


  5. Are their any standard laws/rules of distribution of universal quantifier over conditional and binconditional that can help me solve this?


  6. Also rules for distribution of existential quantifier over conditional and binconditional?



Recently I came across distribution of quantifiers over $vee$ and $wedge$, which gave set theoretic interpretation of them as follows:





  • $((∀x)G(x)∨ (∀x)H(x))→ (∀x)(F(x)∨ G(x))$



    In set theoretic terms,
    if we have that $(f(G) = D ∨ f(H) = D)$, then we have $(f(G) ∪ f(H)) = D$




  • $(∃x)(G(x)∧ H(x))→((∃x)G(x)∧ (∃x)H(x))$



    In set theoretic terms,
    if we have that $(f(G) ∩ f(H)) ≥ 1$, then we have $(f(G) ≥ 1 ∧ f(H) ≥ 1)$




Can we say similar for distribution of quantifiers over conditional and biconditional (just to bring in more clarity)?










share|cite|improve this question















Which of the following propositional logic statements are true and why?




  1. $(∀x(P(x)⟹Q(x)))⟹((∀xP(x))⟹(∀xQ(x)))$

  2. $(∀x(P(x))⟹∀x(Q(x)))⟹(∀x(P(x)⟹Q(x)))$

  3. $(∀x(P(x))⇔(∀x(Q(x))))⟹(∀x(P(x)⇔Q(x)))$

  4. $(∀x(P(x)⇔Q(x)))⟹(∀x(P(x))⇔(∀x(Q(x))))$


  5. Are their any standard laws/rules of distribution of universal quantifier over conditional and binconditional that can help me solve this?


  6. Also rules for distribution of existential quantifier over conditional and binconditional?



Recently I came across distribution of quantifiers over $vee$ and $wedge$, which gave set theoretic interpretation of them as follows:





  • $((∀x)G(x)∨ (∀x)H(x))→ (∀x)(F(x)∨ G(x))$



    In set theoretic terms,
    if we have that $(f(G) = D ∨ f(H) = D)$, then we have $(f(G) ∪ f(H)) = D$




  • $(∃x)(G(x)∧ H(x))→((∃x)G(x)∧ (∃x)H(x))$



    In set theoretic terms,
    if we have that $(f(G) ∩ f(H)) ≥ 1$, then we have $(f(G) ≥ 1 ∧ f(H) ≥ 1)$




Can we say similar for distribution of quantifiers over conditional and biconditional (just to bring in more clarity)?







logic predicate-logic first-order-logic quantifiers






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 1 '16 at 21:50

























asked Jan 1 '16 at 21:20









anir123

855923




855923












  • The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic.
    – Taroccoesbrocco
    Jan 1 '16 at 21:23












  • Do you want me to change the title to something else?
    – anir123
    Jan 1 '16 at 21:25












  • Yes, the title is misleading, as well as the tag "propositional calculus".
    – Taroccoesbrocco
    Jan 1 '16 at 21:27






  • 1




    Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?"
    – anir123
    Jan 1 '16 at 21:31










  • Now the title is ok for me.
    – Taroccoesbrocco
    Jan 1 '16 at 21:33


















  • The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic.
    – Taroccoesbrocco
    Jan 1 '16 at 21:23












  • Do you want me to change the title to something else?
    – anir123
    Jan 1 '16 at 21:25












  • Yes, the title is misleading, as well as the tag "propositional calculus".
    – Taroccoesbrocco
    Jan 1 '16 at 21:27






  • 1




    Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?"
    – anir123
    Jan 1 '16 at 21:31










  • Now the title is ok for me.
    – Taroccoesbrocco
    Jan 1 '16 at 21:33
















The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic.
– Taroccoesbrocco
Jan 1 '16 at 21:23






The above formulas are not propositional, since they use quantifiers. They are just formulas of first-order logic.
– Taroccoesbrocco
Jan 1 '16 at 21:23














Do you want me to change the title to something else?
– anir123
Jan 1 '16 at 21:25






Do you want me to change the title to something else?
– anir123
Jan 1 '16 at 21:25














Yes, the title is misleading, as well as the tag "propositional calculus".
– Taroccoesbrocco
Jan 1 '16 at 21:27




Yes, the title is misleading, as well as the tag "propositional calculus".
– Taroccoesbrocco
Jan 1 '16 at 21:27




1




1




Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?"
– anir123
Jan 1 '16 at 21:31




Does the new one makes sense "Rules of distribution of quantifiers over conditional and biconditional makes sense?"
– anir123
Jan 1 '16 at 21:31












Now the title is ok for me.
– Taroccoesbrocco
Jan 1 '16 at 21:33




Now the title is ok for me.
– Taroccoesbrocco
Jan 1 '16 at 21:33










1 Answer
1






active

oldest

votes


















0














Formulas (1) and (4) are valid, i.e. they are true in every first-order $mathcal{L}$-structure.



Formulas (2) and (3) are not valid, i.e. there exists a $mathcal{L}$-structure in which they are not true. For instance, take the $mathcal{L}$-structure $mathcal{N}$ whose domain is $mathbb{N}$ and whose interpretation of $P$ is $2mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $mathbb{N} smallsetminus 2mathbb{N}$ (the set of odd natural numbers). You have that the formula $forall xP(x) Rightarrow forall x Q(x)$ is vacuously true in $mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $forall x(P(x) Rightarrow Q(x))$ is false in $mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $mathcal{N}$. Similarly for the formula (3), since $A Leftrightarrow B$ is equivalent to $(A Rightarrow B) land (B Rightarrow A)$.



In general, when one talks about distributivity of something over something else (for instance, distributivity of $land$ over $lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $forall xP(x) Rightarrow forall x Q(x)$ is not logically equivalent to the formula $forall x(P(x) Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $forall xP(x) Leftrightarrow forall x Q(x)$ is not logically equivalent to formula $forall x(P(x) Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).






share|cite|improve this answer























  • Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
    – anir123
    Jan 2 '16 at 9:10










  • ...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
    – anir123
    Jan 2 '16 at 9:11












  • ...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
    – anir123
    Jan 2 '16 at 9:11












  • @PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
    – Taroccoesbrocco
    Jan 2 '16 at 10:14










  • any comment about the same but involving existential quantifier?
    – anir123
    Jan 2 '16 at 12:50











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1 Answer
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1 Answer
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oldest

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Formulas (1) and (4) are valid, i.e. they are true in every first-order $mathcal{L}$-structure.



Formulas (2) and (3) are not valid, i.e. there exists a $mathcal{L}$-structure in which they are not true. For instance, take the $mathcal{L}$-structure $mathcal{N}$ whose domain is $mathbb{N}$ and whose interpretation of $P$ is $2mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $mathbb{N} smallsetminus 2mathbb{N}$ (the set of odd natural numbers). You have that the formula $forall xP(x) Rightarrow forall x Q(x)$ is vacuously true in $mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $forall x(P(x) Rightarrow Q(x))$ is false in $mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $mathcal{N}$. Similarly for the formula (3), since $A Leftrightarrow B$ is equivalent to $(A Rightarrow B) land (B Rightarrow A)$.



In general, when one talks about distributivity of something over something else (for instance, distributivity of $land$ over $lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $forall xP(x) Rightarrow forall x Q(x)$ is not logically equivalent to the formula $forall x(P(x) Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $forall xP(x) Leftrightarrow forall x Q(x)$ is not logically equivalent to formula $forall x(P(x) Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).






share|cite|improve this answer























  • Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
    – anir123
    Jan 2 '16 at 9:10










  • ...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
    – anir123
    Jan 2 '16 at 9:11












  • ...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
    – anir123
    Jan 2 '16 at 9:11












  • @PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
    – Taroccoesbrocco
    Jan 2 '16 at 10:14










  • any comment about the same but involving existential quantifier?
    – anir123
    Jan 2 '16 at 12:50
















0














Formulas (1) and (4) are valid, i.e. they are true in every first-order $mathcal{L}$-structure.



Formulas (2) and (3) are not valid, i.e. there exists a $mathcal{L}$-structure in which they are not true. For instance, take the $mathcal{L}$-structure $mathcal{N}$ whose domain is $mathbb{N}$ and whose interpretation of $P$ is $2mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $mathbb{N} smallsetminus 2mathbb{N}$ (the set of odd natural numbers). You have that the formula $forall xP(x) Rightarrow forall x Q(x)$ is vacuously true in $mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $forall x(P(x) Rightarrow Q(x))$ is false in $mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $mathcal{N}$. Similarly for the formula (3), since $A Leftrightarrow B$ is equivalent to $(A Rightarrow B) land (B Rightarrow A)$.



In general, when one talks about distributivity of something over something else (for instance, distributivity of $land$ over $lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $forall xP(x) Rightarrow forall x Q(x)$ is not logically equivalent to the formula $forall x(P(x) Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $forall xP(x) Leftrightarrow forall x Q(x)$ is not logically equivalent to formula $forall x(P(x) Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).






share|cite|improve this answer























  • Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
    – anir123
    Jan 2 '16 at 9:10










  • ...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
    – anir123
    Jan 2 '16 at 9:11












  • ...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
    – anir123
    Jan 2 '16 at 9:11












  • @PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
    – Taroccoesbrocco
    Jan 2 '16 at 10:14










  • any comment about the same but involving existential quantifier?
    – anir123
    Jan 2 '16 at 12:50














0












0








0






Formulas (1) and (4) are valid, i.e. they are true in every first-order $mathcal{L}$-structure.



Formulas (2) and (3) are not valid, i.e. there exists a $mathcal{L}$-structure in which they are not true. For instance, take the $mathcal{L}$-structure $mathcal{N}$ whose domain is $mathbb{N}$ and whose interpretation of $P$ is $2mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $mathbb{N} smallsetminus 2mathbb{N}$ (the set of odd natural numbers). You have that the formula $forall xP(x) Rightarrow forall x Q(x)$ is vacuously true in $mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $forall x(P(x) Rightarrow Q(x))$ is false in $mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $mathcal{N}$. Similarly for the formula (3), since $A Leftrightarrow B$ is equivalent to $(A Rightarrow B) land (B Rightarrow A)$.



In general, when one talks about distributivity of something over something else (for instance, distributivity of $land$ over $lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $forall xP(x) Rightarrow forall x Q(x)$ is not logically equivalent to the formula $forall x(P(x) Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $forall xP(x) Leftrightarrow forall x Q(x)$ is not logically equivalent to formula $forall x(P(x) Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).






share|cite|improve this answer














Formulas (1) and (4) are valid, i.e. they are true in every first-order $mathcal{L}$-structure.



Formulas (2) and (3) are not valid, i.e. there exists a $mathcal{L}$-structure in which they are not true. For instance, take the $mathcal{L}$-structure $mathcal{N}$ whose domain is $mathbb{N}$ and whose interpretation of $P$ is $2mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $mathbb{N} smallsetminus 2mathbb{N}$ (the set of odd natural numbers). You have that the formula $forall xP(x) Rightarrow forall x Q(x)$ is vacuously true in $mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $forall x(P(x) Rightarrow Q(x))$ is false in $mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $mathcal{N}$. Similarly for the formula (3), since $A Leftrightarrow B$ is equivalent to $(A Rightarrow B) land (B Rightarrow A)$.



In general, when one talks about distributivity of something over something else (for instance, distributivity of $land$ over $lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $forall xP(x) Rightarrow forall x Q(x)$ is not logically equivalent to the formula $forall x(P(x) Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $forall xP(x) Leftrightarrow forall x Q(x)$ is not logically equivalent to formula $forall x(P(x) Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 '16 at 16:09

























answered Jan 1 '16 at 21:42









Taroccoesbrocco

4,93761838




4,93761838












  • Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
    – anir123
    Jan 2 '16 at 9:10










  • ...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
    – anir123
    Jan 2 '16 at 9:11












  • ...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
    – anir123
    Jan 2 '16 at 9:11












  • @PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
    – Taroccoesbrocco
    Jan 2 '16 at 10:14










  • any comment about the same but involving existential quantifier?
    – anir123
    Jan 2 '16 at 12:50


















  • Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
    – anir123
    Jan 2 '16 at 9:10










  • ...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
    – anir123
    Jan 2 '16 at 9:11












  • ...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
    – anir123
    Jan 2 '16 at 9:11












  • @PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
    – Taroccoesbrocco
    Jan 2 '16 at 10:14










  • any comment about the same but involving existential quantifier?
    – anir123
    Jan 2 '16 at 12:50
















Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
– anir123
Jan 2 '16 at 9:10




Well I have not taken a dedicated course on mathematical strucutres or read any books specifically on mathematical structures. So I am finding it difficult to deal with $L$-structure thingy...
– anir123
Jan 2 '16 at 9:10












...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
– anir123
Jan 2 '16 at 9:11






...However,if interprete $P(x)$ as "x is a boy" & "$Q(x)$ is clever". Then, $(∀x(P(x)⇔Q(x)))$ means "If any $x$ is a boy then he is clever & vice versa". This definitely implies $(∀x(P(x))⇔(∀x(Q(x))))$, "If all $x$ are boys then all are clever & vice versa". But converse does not seem valid as $(∀x(P(x))⇔(∀x(Q(x))))$ puts restriction: "if all $x$ are boys". $(∀x(P(x)⇔Q(x)))$ requires that "for any $x$, if $x$ is a boy", but does not require that "all $x$ are boys", so there may be a person who is not boy, but still $(∀x(P(x)⇔Q(x)))$ will be valid but $(∀x(P(x))⇔(∀x(Q(x))))$ will not be...
– anir123
Jan 2 '16 at 9:11














...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
– anir123
Jan 2 '16 at 9:11






...So 3 seems invalid to me, while 4 seems valid. Also what about distributing existential quantifier?
– anir123
Jan 2 '16 at 9:11














@PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
– Taroccoesbrocco
Jan 2 '16 at 10:14




@PardonMeForMySuperPoorMaths: Sorry, there was a typo in my answer, now I fixed it. Your formula (4) is the "biconditional version" of your formula (1), and your formula (3) is the "biconditional version" of your formula (2).
– Taroccoesbrocco
Jan 2 '16 at 10:14












any comment about the same but involving existential quantifier?
– anir123
Jan 2 '16 at 12:50




any comment about the same but involving existential quantifier?
– anir123
Jan 2 '16 at 12:50


















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