how to prove ∃x(∃yA(y) → A(x)) is valid in classical logic











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∃x(∃yA(y) → A(x))
i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
can anyone help me please ?










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    up vote
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    down vote

    favorite
    1












    ∃x(∃yA(y) → A(x))
    i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
    can anyone help me please ?










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      ∃x(∃yA(y) → A(x))
      i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
      can anyone help me please ?










      share|cite|improve this question















      ∃x(∃yA(y) → A(x))
      i know how to prove it by natural deduction. but it requires to use model theoretical approach. so i'm not sure how we can prove the validity.
      can anyone help me please ?







      logic predicate-logic






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      share|cite|improve this question













      share|cite|improve this question




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      edited Nov 21 at 14:58









      Mauro ALLEGRANZA

      63.9k448110




      63.9k448110










      asked Nov 21 at 14:18









      Norman

      227




      227






















          2 Answers
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          up vote
          2
          down vote



          accepted










          Let $mathcal M$ a structure whatever, with domain $D$.



          Two cases :



          (i) $∃yA(y)$ is False.



          Thus $∃yA(y) to A(x)$ is True.



          For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.



          And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.



          (ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.



          This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.






          share|cite|improve this answer























          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01


















          up vote
          1
          down vote













          Try a proof by contradiction:



          So, suppose there is some structure that makes this sentence false.



          That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.



          By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.



          By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.



          But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.



          So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.



          Hence, the sentence is valid.






          share|cite|improve this answer























          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01










          • @Norman You're welcome :)
            – Bram28
            Nov 21 at 17:37











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          Let $mathcal M$ a structure whatever, with domain $D$.



          Two cases :



          (i) $∃yA(y)$ is False.



          Thus $∃yA(y) to A(x)$ is True.



          For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.



          And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.



          (ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.



          This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.






          share|cite|improve this answer























          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01















          up vote
          2
          down vote



          accepted










          Let $mathcal M$ a structure whatever, with domain $D$.



          Two cases :



          (i) $∃yA(y)$ is False.



          Thus $∃yA(y) to A(x)$ is True.



          For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.



          And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.



          (ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.



          This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.






          share|cite|improve this answer























          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Let $mathcal M$ a structure whatever, with domain $D$.



          Two cases :



          (i) $∃yA(y)$ is False.



          Thus $∃yA(y) to A(x)$ is True.



          For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.



          And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.



          (ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.



          This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.






          share|cite|improve this answer














          Let $mathcal M$ a structure whatever, with domain $D$.



          Two cases :



          (i) $∃yA(y)$ is False.



          Thus $∃yA(y) to A(x)$ is True.



          For an assignment $s$ such that $s(x)=a$, for $a in D$ whatever, we have $mathcal M vDash (∃yA(y) to A(x))[s]$.



          And this means that $mathcal M vDash exists x (∃yA(y) to A(x))$.



          (ii) $∃yA(y)$ is True, i.e. there is an object $a in D$ such that $A^{mathcal M}(a)$ holds.



          This implies that $mathcal M vDash (∃yA(y) to A(x))[s]$, for an assignment $s$ such that $s(x)=a$, and thus : $mathcal M vDash exists x (∃yA(y) to A(x))$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 14:43

























          answered Nov 21 at 14:37









          Mauro ALLEGRANZA

          63.9k448110




          63.9k448110












          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01


















          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01
















          Thank you so much. i really appreciate it.
          – Norman
          Nov 21 at 16:01




          Thank you so much. i really appreciate it.
          – Norman
          Nov 21 at 16:01










          up vote
          1
          down vote













          Try a proof by contradiction:



          So, suppose there is some structure that makes this sentence false.



          That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.



          By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.



          By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.



          But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.



          So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.



          Hence, the sentence is valid.






          share|cite|improve this answer























          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01










          • @Norman You're welcome :)
            – Bram28
            Nov 21 at 17:37















          up vote
          1
          down vote













          Try a proof by contradiction:



          So, suppose there is some structure that makes this sentence false.



          That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.



          By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.



          By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.



          But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.



          So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.



          Hence, the sentence is valid.






          share|cite|improve this answer























          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01










          • @Norman You're welcome :)
            – Bram28
            Nov 21 at 17:37













          up vote
          1
          down vote










          up vote
          1
          down vote









          Try a proof by contradiction:



          So, suppose there is some structure that makes this sentence false.



          That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.



          By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.



          By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.



          But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.



          So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.



          Hence, the sentence is valid.






          share|cite|improve this answer














          Try a proof by contradiction:



          So, suppose there is some structure that makes this sentence false.



          That means that there is not some object $x$ in the domain for which it holds that if there is something which has property $A$, then $x$ has property $A$.



          By pure logic that means that for all objects $x$ in the domain it does not hold that if there is something which has property $A$, then $x$ has property $A$.



          By the semantics of the material implication, this then means that for all objects $x$ in the domain it holds that there is something with property $A$, and that $x$ does not have property $A$.



          But that means that there is something with property $A$ as well as that nothing has property $A$, and we have our contradiction.



          So, there is no structure that makes the sentence false, and hence all structures will set this sentence to true.



          Hence, the sentence is valid.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 21 at 14:54

























          answered Nov 21 at 14:31









          Bram28

          58.8k44185




          58.8k44185












          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01










          • @Norman You're welcome :)
            – Bram28
            Nov 21 at 17:37


















          • Thank you so much. i really appreciate it.
            – Norman
            Nov 21 at 16:01










          • @Norman You're welcome :)
            – Bram28
            Nov 21 at 17:37
















          Thank you so much. i really appreciate it.
          – Norman
          Nov 21 at 16:01




          Thank you so much. i really appreciate it.
          – Norman
          Nov 21 at 16:01












          @Norman You're welcome :)
          – Bram28
          Nov 21 at 17:37




          @Norman You're welcome :)
          – Bram28
          Nov 21 at 17:37


















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