Difficulty with understanding the semantics of the following equivalences











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I know that the rules for removing quantifiers from the antecedent of a conditional are as follows:
(∀x)φ(x) → ψ ⇔ (∃x)(φ(x) → ψ) and (∃x)φ(x) → ψ ⇔ (∀x)(φ(x) → ψ), provided that x is not free in ψ.
I also know that the rules for removing quantifiers from the consequent are as follows:
φ → (∀x)ψ(x) ⇔ (∀x)(φ → ψ(x)) and φ → (∃x)ψ(x) ⇔ (∃x)(φ → ψ(x)), provided that x is not free in φ.



Now, I understand how to derive these equivalences syntactically. However, I can't seem to grasp how the first two can be semantically equivalent.

For example, how can two sentences such as "If all x are wet, then it's raining" and "There exists an x such that if x is wet, then it's raining" can be equivalent?

Moreover, provided that x is not free in ψ, isn't (∀x)φ(x) → ψ equivalent to (∀x)(φ(x) → ψ)? Which would result in the absurd equivalence: (∀x)(φ(x) → ψ) ⇔ (∃x)(φ(x) → ψ).



I think that my difficulty here arises from a misunderstanding of the significance of the parentheses in the aforementioned equivalences. Please provide examples in your answer if you can.










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    up vote
    7
    down vote

    favorite
    1












    I know that the rules for removing quantifiers from the antecedent of a conditional are as follows:
    (∀x)φ(x) → ψ ⇔ (∃x)(φ(x) → ψ) and (∃x)φ(x) → ψ ⇔ (∀x)(φ(x) → ψ), provided that x is not free in ψ.
    I also know that the rules for removing quantifiers from the consequent are as follows:
    φ → (∀x)ψ(x) ⇔ (∀x)(φ → ψ(x)) and φ → (∃x)ψ(x) ⇔ (∃x)(φ → ψ(x)), provided that x is not free in φ.



    Now, I understand how to derive these equivalences syntactically. However, I can't seem to grasp how the first two can be semantically equivalent.

    For example, how can two sentences such as "If all x are wet, then it's raining" and "There exists an x such that if x is wet, then it's raining" can be equivalent?

    Moreover, provided that x is not free in ψ, isn't (∀x)φ(x) → ψ equivalent to (∀x)(φ(x) → ψ)? Which would result in the absurd equivalence: (∀x)(φ(x) → ψ) ⇔ (∃x)(φ(x) → ψ).



    I think that my difficulty here arises from a misunderstanding of the significance of the parentheses in the aforementioned equivalences. Please provide examples in your answer if you can.










    share|improve this question


























      up vote
      7
      down vote

      favorite
      1









      up vote
      7
      down vote

      favorite
      1






      1





      I know that the rules for removing quantifiers from the antecedent of a conditional are as follows:
      (∀x)φ(x) → ψ ⇔ (∃x)(φ(x) → ψ) and (∃x)φ(x) → ψ ⇔ (∀x)(φ(x) → ψ), provided that x is not free in ψ.
      I also know that the rules for removing quantifiers from the consequent are as follows:
      φ → (∀x)ψ(x) ⇔ (∀x)(φ → ψ(x)) and φ → (∃x)ψ(x) ⇔ (∃x)(φ → ψ(x)), provided that x is not free in φ.



      Now, I understand how to derive these equivalences syntactically. However, I can't seem to grasp how the first two can be semantically equivalent.

      For example, how can two sentences such as "If all x are wet, then it's raining" and "There exists an x such that if x is wet, then it's raining" can be equivalent?

      Moreover, provided that x is not free in ψ, isn't (∀x)φ(x) → ψ equivalent to (∀x)(φ(x) → ψ)? Which would result in the absurd equivalence: (∀x)(φ(x) → ψ) ⇔ (∃x)(φ(x) → ψ).



      I think that my difficulty here arises from a misunderstanding of the significance of the parentheses in the aforementioned equivalences. Please provide examples in your answer if you can.










      share|improve this question















      I know that the rules for removing quantifiers from the antecedent of a conditional are as follows:
      (∀x)φ(x) → ψ ⇔ (∃x)(φ(x) → ψ) and (∃x)φ(x) → ψ ⇔ (∀x)(φ(x) → ψ), provided that x is not free in ψ.
      I also know that the rules for removing quantifiers from the consequent are as follows:
      φ → (∀x)ψ(x) ⇔ (∀x)(φ → ψ(x)) and φ → (∃x)ψ(x) ⇔ (∃x)(φ → ψ(x)), provided that x is not free in φ.



      Now, I understand how to derive these equivalences syntactically. However, I can't seem to grasp how the first two can be semantically equivalent.

      For example, how can two sentences such as "If all x are wet, then it's raining" and "There exists an x such that if x is wet, then it's raining" can be equivalent?

      Moreover, provided that x is not free in ψ, isn't (∀x)φ(x) → ψ equivalent to (∀x)(φ(x) → ψ)? Which would result in the absurd equivalence: (∀x)(φ(x) → ψ) ⇔ (∃x)(φ(x) → ψ).



      I think that my difficulty here arises from a misunderstanding of the significance of the parentheses in the aforementioned equivalences. Please provide examples in your answer if you can.







      logic






      share|improve this question















      share|improve this question













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      share|improve this question








      edited Nov 25 at 10:16

























      asked Nov 25 at 9:56









      Ofek Aman

      384




      384






















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          how can two sentences such as "(If all x are wet, then it's raining)" and "There exists an x such that (if x is wet, then it's raining)" can be equivalent?




          A simple approach to this equivalence, that holds only in classical logic, is to exploit the so-called Material implication rule :




          (P → Q) ≡ (¬ P ∨ Q).




          Thus, we have that : "(If all x are wet, then it's raining)" is equivalent to : "Either (not all x are wet) or it's raining", i.e "Either (there exists an x which is not wet) or it's raining".



          Now we have to use the fact that the existential quantifier "distribute" over or and the fact that "it's is raining" is equivalent to "there is an x such that it's raining", due to the key-fact that x is not free in "it's raining".



          Thus, the last step is to move the existential quantifier in front of the formula to get :




          "There exists an x such that (either x is not wet or it's raining)."






          We can show the semantical equivalence of :




          (∀x βx → α) ↔ ∃x(βx → α)




          this way.



          Let M a structure whatever, with domain D.



          Two cases :



          (i) ∀x βx is False. In this case (∀x βx → α) is True.



          This means that there is an a in D such that βa is False.



          Thus, (βa → α) is True, which means that (βx → α) is True for some x.



          And thus, also ∃x(βx → α) is True.



          (ii) ∀x βx is True.



          Two subcases :



          (iia) α is True. In this case (∀x βx → α) is True.



          But if α is True, also (βx → α), irrespective of the value of x, and thus also ∃x(βx → α) is True.



          (iib) α is False. In this case (∀x βx → α) is False.



          The fact that ∀x βx is True imples that βa is True, for every a in D.



          Thus (βa → α) is False, for every a in D, which means that ∃x(βx → α) is False.






          share|improve this answer



















          • 1




            Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
            – Ofek Aman
            Nov 25 at 11:15










          • @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
            – Mauro ALLEGRANZA
            Nov 25 at 11:19












          • Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
            – Ofek Aman
            Nov 25 at 11:23











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          up vote
          7
          down vote



          accepted











          how can two sentences such as "(If all x are wet, then it's raining)" and "There exists an x such that (if x is wet, then it's raining)" can be equivalent?




          A simple approach to this equivalence, that holds only in classical logic, is to exploit the so-called Material implication rule :




          (P → Q) ≡ (¬ P ∨ Q).




          Thus, we have that : "(If all x are wet, then it's raining)" is equivalent to : "Either (not all x are wet) or it's raining", i.e "Either (there exists an x which is not wet) or it's raining".



          Now we have to use the fact that the existential quantifier "distribute" over or and the fact that "it's is raining" is equivalent to "there is an x such that it's raining", due to the key-fact that x is not free in "it's raining".



          Thus, the last step is to move the existential quantifier in front of the formula to get :




          "There exists an x such that (either x is not wet or it's raining)."






          We can show the semantical equivalence of :




          (∀x βx → α) ↔ ∃x(βx → α)




          this way.



          Let M a structure whatever, with domain D.



          Two cases :



          (i) ∀x βx is False. In this case (∀x βx → α) is True.



          This means that there is an a in D such that βa is False.



          Thus, (βa → α) is True, which means that (βx → α) is True for some x.



          And thus, also ∃x(βx → α) is True.



          (ii) ∀x βx is True.



          Two subcases :



          (iia) α is True. In this case (∀x βx → α) is True.



          But if α is True, also (βx → α), irrespective of the value of x, and thus also ∃x(βx → α) is True.



          (iib) α is False. In this case (∀x βx → α) is False.



          The fact that ∀x βx is True imples that βa is True, for every a in D.



          Thus (βa → α) is False, for every a in D, which means that ∃x(βx → α) is False.






          share|improve this answer



















          • 1




            Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
            – Ofek Aman
            Nov 25 at 11:15










          • @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
            – Mauro ALLEGRANZA
            Nov 25 at 11:19












          • Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
            – Ofek Aman
            Nov 25 at 11:23















          up vote
          7
          down vote



          accepted











          how can two sentences such as "(If all x are wet, then it's raining)" and "There exists an x such that (if x is wet, then it's raining)" can be equivalent?




          A simple approach to this equivalence, that holds only in classical logic, is to exploit the so-called Material implication rule :




          (P → Q) ≡ (¬ P ∨ Q).




          Thus, we have that : "(If all x are wet, then it's raining)" is equivalent to : "Either (not all x are wet) or it's raining", i.e "Either (there exists an x which is not wet) or it's raining".



          Now we have to use the fact that the existential quantifier "distribute" over or and the fact that "it's is raining" is equivalent to "there is an x such that it's raining", due to the key-fact that x is not free in "it's raining".



          Thus, the last step is to move the existential quantifier in front of the formula to get :




          "There exists an x such that (either x is not wet or it's raining)."






          We can show the semantical equivalence of :




          (∀x βx → α) ↔ ∃x(βx → α)




          this way.



          Let M a structure whatever, with domain D.



          Two cases :



          (i) ∀x βx is False. In this case (∀x βx → α) is True.



          This means that there is an a in D such that βa is False.



          Thus, (βa → α) is True, which means that (βx → α) is True for some x.



          And thus, also ∃x(βx → α) is True.



          (ii) ∀x βx is True.



          Two subcases :



          (iia) α is True. In this case (∀x βx → α) is True.



          But if α is True, also (βx → α), irrespective of the value of x, and thus also ∃x(βx → α) is True.



          (iib) α is False. In this case (∀x βx → α) is False.



          The fact that ∀x βx is True imples that βa is True, for every a in D.



          Thus (βa → α) is False, for every a in D, which means that ∃x(βx → α) is False.






          share|improve this answer



















          • 1




            Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
            – Ofek Aman
            Nov 25 at 11:15










          • @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
            – Mauro ALLEGRANZA
            Nov 25 at 11:19












          • Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
            – Ofek Aman
            Nov 25 at 11:23













          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted







          how can two sentences such as "(If all x are wet, then it's raining)" and "There exists an x such that (if x is wet, then it's raining)" can be equivalent?




          A simple approach to this equivalence, that holds only in classical logic, is to exploit the so-called Material implication rule :




          (P → Q) ≡ (¬ P ∨ Q).




          Thus, we have that : "(If all x are wet, then it's raining)" is equivalent to : "Either (not all x are wet) or it's raining", i.e "Either (there exists an x which is not wet) or it's raining".



          Now we have to use the fact that the existential quantifier "distribute" over or and the fact that "it's is raining" is equivalent to "there is an x such that it's raining", due to the key-fact that x is not free in "it's raining".



          Thus, the last step is to move the existential quantifier in front of the formula to get :




          "There exists an x such that (either x is not wet or it's raining)."






          We can show the semantical equivalence of :




          (∀x βx → α) ↔ ∃x(βx → α)




          this way.



          Let M a structure whatever, with domain D.



          Two cases :



          (i) ∀x βx is False. In this case (∀x βx → α) is True.



          This means that there is an a in D such that βa is False.



          Thus, (βa → α) is True, which means that (βx → α) is True for some x.



          And thus, also ∃x(βx → α) is True.



          (ii) ∀x βx is True.



          Two subcases :



          (iia) α is True. In this case (∀x βx → α) is True.



          But if α is True, also (βx → α), irrespective of the value of x, and thus also ∃x(βx → α) is True.



          (iib) α is False. In this case (∀x βx → α) is False.



          The fact that ∀x βx is True imples that βa is True, for every a in D.



          Thus (βa → α) is False, for every a in D, which means that ∃x(βx → α) is False.






          share|improve this answer















          how can two sentences such as "(If all x are wet, then it's raining)" and "There exists an x such that (if x is wet, then it's raining)" can be equivalent?




          A simple approach to this equivalence, that holds only in classical logic, is to exploit the so-called Material implication rule :




          (P → Q) ≡ (¬ P ∨ Q).




          Thus, we have that : "(If all x are wet, then it's raining)" is equivalent to : "Either (not all x are wet) or it's raining", i.e "Either (there exists an x which is not wet) or it's raining".



          Now we have to use the fact that the existential quantifier "distribute" over or and the fact that "it's is raining" is equivalent to "there is an x such that it's raining", due to the key-fact that x is not free in "it's raining".



          Thus, the last step is to move the existential quantifier in front of the formula to get :




          "There exists an x such that (either x is not wet or it's raining)."






          We can show the semantical equivalence of :




          (∀x βx → α) ↔ ∃x(βx → α)




          this way.



          Let M a structure whatever, with domain D.



          Two cases :



          (i) ∀x βx is False. In this case (∀x βx → α) is True.



          This means that there is an a in D such that βa is False.



          Thus, (βa → α) is True, which means that (βx → α) is True for some x.



          And thus, also ∃x(βx → α) is True.



          (ii) ∀x βx is True.



          Two subcases :



          (iia) α is True. In this case (∀x βx → α) is True.



          But if α is True, also (βx → α), irrespective of the value of x, and thus also ∃x(βx → α) is True.



          (iib) α is False. In this case (∀x βx → α) is False.



          The fact that ∀x βx is True imples that βa is True, for every a in D.



          Thus (βa → α) is False, for every a in D, which means that ∃x(βx → α) is False.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 25 at 11:18

























          answered Nov 25 at 10:11









          Mauro ALLEGRANZA

          27k21860




          27k21860








          • 1




            Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
            – Ofek Aman
            Nov 25 at 11:15










          • @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
            – Mauro ALLEGRANZA
            Nov 25 at 11:19












          • Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
            – Ofek Aman
            Nov 25 at 11:23














          • 1




            Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
            – Ofek Aman
            Nov 25 at 11:15










          • @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
            – Mauro ALLEGRANZA
            Nov 25 at 11:19












          • Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
            – Ofek Aman
            Nov 25 at 11:23








          1




          1




          Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
          – Ofek Aman
          Nov 25 at 11:15




          Thank you for clarifying the problem. I understand how the universal conditional implies the existential so no problem here. However, I still don't get how the existential conditional also implies the universal.
          – Ofek Aman
          Nov 25 at 11:15












          @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
          – Mauro ALLEGRANZA
          Nov 25 at 11:19






          @OfekAman - not clear... the two are equivalent : when (∀x βx → α) is True, also ∃x(βx → α) is True. When (∀x βx → α) is False, also ∃x(βx → α) is False.
          – Mauro ALLEGRANZA
          Nov 25 at 11:19














          Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
          – Ofek Aman
          Nov 25 at 11:23




          Oh, I see it now. The two expressions always get yhe same truth value and therefore are equivalent. Thank you!
          – Ofek Aman
          Nov 25 at 11:23


















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