Write in Logic: If professors are unhappy all students fail their exams












1












$begingroup$


I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:



∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]



However, the answer of my teacher is:



∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))



Can someone helps me?






logic first-order-logic







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  • $begingroup$
    The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 13 '18 at 9:28
















1












$begingroup$


I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:



∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]



However, the answer of my teacher is:



∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))



Can someone helps me?






logic first-order-logic







share|cite|improve this question









$endgroup$












  • $begingroup$
    The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 13 '18 at 9:28














1












1








1





$begingroup$


I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:



∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]



However, the answer of my teacher is:



∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))



Can someone helps me?






logic first-order-logic







share|cite|improve this question









$endgroup$




I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:



∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]



However, the answer of my teacher is:



∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))



Can someone helps me?






logic first-order-logic




logic first-order-logic






share|cite|improve this question













share|cite|improve this question











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asked Dec 13 '18 at 9:22









GoldGold

61




61












  • $begingroup$
    The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 13 '18 at 9:28


















  • $begingroup$
    The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 13 '18 at 9:28
















$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28




$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28










2 Answers
2






active

oldest

votes


















0












$begingroup$

They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:



$forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$



and



$forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$



Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:



$psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$



Here, the formula $psi$ cannot include any free variables $x$



Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:



$forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$



Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:



$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$



Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:



      $forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$



      and



      $forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$



      Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:



      $psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$



      Here, the formula $psi$ cannot include any free variables $x$



      Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:



      $forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$



      Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:



      $P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$



      Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:



        $forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$



        and



        $forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$



        Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:



        $psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$



        Here, the formula $psi$ cannot include any free variables $x$



        Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:



        $forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$



        Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:



        $P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$



        Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:



          $forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$



          and



          $forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$



          Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:



          $psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$



          Here, the formula $psi$ cannot include any free variables $x$



          Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:



          $forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$



          Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:



          $P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$



          Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.






          share|cite|improve this answer









          $endgroup$



          They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:



          $forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$



          and



          $forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$



          Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:



          $psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$



          Here, the formula $psi$ cannot include any free variables $x$



          Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:



          $forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$



          Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:



          $P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$



          Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 13:12









          Bram28Bram28

          64.2k44793




          64.2k44793























              0












              $begingroup$

              You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.






                  share|cite|improve this answer









                  $endgroup$



                  You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 14:52









                  Jorge AdrianoJorge Adriano

                  59146




                  59146






























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