Predicate logic - logically imply












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$begingroup$




  • Consider the following predicate formulas.



    $F1: forall x exists y ( P(x) to Q(y) ).$



    $F2: exists x forall y ( P(x) to Q(y) ).$



    $F3: forall x P(x) to exists y Q(y).$



    $F4: exists x P(x) to forall y Q(y).$



    Answer the following questions with brief justification.



    (a) Does $F1$ logically imply $F2$?



    (b) Does $F1$ logically imply $F3$?



    (c) Does $F1$ logically imply $F4$?



    (d) Does $F2$ logically imply $F1$?



    (e) Does $F2$ logically imply $F3$?



    (f) Does $F2$ logically imply $F4$?



    (g) Does $F3$ logically imply $F1$?



    (h) Does $F3$ logically imply $F2$?



    (i) Does $F3$ logically imply $F4$?



    (j) Does $F4$ logically imply $F1$?



    (k) Does $F4$ logically imply $F2$?



    (l) Does $F4$ logically imply $F3$?






Logically implies really confuses me. Attempt:



Writing it in English first:



F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$



F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$



F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$



F4: $x$ never satisfies $P$ or $y$ always satisfies Q



By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$



So:



(a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.



(b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped



Apparently $b$ is true I don't get it though.



according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.



then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.










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$endgroup$

















    0












    $begingroup$




    • Consider the following predicate formulas.



      $F1: forall x exists y ( P(x) to Q(y) ).$



      $F2: exists x forall y ( P(x) to Q(y) ).$



      $F3: forall x P(x) to exists y Q(y).$



      $F4: exists x P(x) to forall y Q(y).$



      Answer the following questions with brief justification.



      (a) Does $F1$ logically imply $F2$?



      (b) Does $F1$ logically imply $F3$?



      (c) Does $F1$ logically imply $F4$?



      (d) Does $F2$ logically imply $F1$?



      (e) Does $F2$ logically imply $F3$?



      (f) Does $F2$ logically imply $F4$?



      (g) Does $F3$ logically imply $F1$?



      (h) Does $F3$ logically imply $F2$?



      (i) Does $F3$ logically imply $F4$?



      (j) Does $F4$ logically imply $F1$?



      (k) Does $F4$ logically imply $F2$?



      (l) Does $F4$ logically imply $F3$?






    Logically implies really confuses me. Attempt:



    Writing it in English first:



    F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$



    F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$



    F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$



    F4: $x$ never satisfies $P$ or $y$ always satisfies Q



    By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$



    So:



    (a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.



    (b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped



    Apparently $b$ is true I don't get it though.



    according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.



    then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$




      • Consider the following predicate formulas.



        $F1: forall x exists y ( P(x) to Q(y) ).$



        $F2: exists x forall y ( P(x) to Q(y) ).$



        $F3: forall x P(x) to exists y Q(y).$



        $F4: exists x P(x) to forall y Q(y).$



        Answer the following questions with brief justification.



        (a) Does $F1$ logically imply $F2$?



        (b) Does $F1$ logically imply $F3$?



        (c) Does $F1$ logically imply $F4$?



        (d) Does $F2$ logically imply $F1$?



        (e) Does $F2$ logically imply $F3$?



        (f) Does $F2$ logically imply $F4$?



        (g) Does $F3$ logically imply $F1$?



        (h) Does $F3$ logically imply $F2$?



        (i) Does $F3$ logically imply $F4$?



        (j) Does $F4$ logically imply $F1$?



        (k) Does $F4$ logically imply $F2$?



        (l) Does $F4$ logically imply $F3$?






      Logically implies really confuses me. Attempt:



      Writing it in English first:



      F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$



      F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$



      F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$



      F4: $x$ never satisfies $P$ or $y$ always satisfies Q



      By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$



      So:



      (a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.



      (b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped



      Apparently $b$ is true I don't get it though.



      according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.



      then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.










      share|cite|improve this question









      $endgroup$






      • Consider the following predicate formulas.



        $F1: forall x exists y ( P(x) to Q(y) ).$



        $F2: exists x forall y ( P(x) to Q(y) ).$



        $F3: forall x P(x) to exists y Q(y).$



        $F4: exists x P(x) to forall y Q(y).$



        Answer the following questions with brief justification.



        (a) Does $F1$ logically imply $F2$?



        (b) Does $F1$ logically imply $F3$?



        (c) Does $F1$ logically imply $F4$?



        (d) Does $F2$ logically imply $F1$?



        (e) Does $F2$ logically imply $F3$?



        (f) Does $F2$ logically imply $F4$?



        (g) Does $F3$ logically imply $F1$?



        (h) Does $F3$ logically imply $F2$?



        (i) Does $F3$ logically imply $F4$?



        (j) Does $F4$ logically imply $F1$?



        (k) Does $F4$ logically imply $F2$?



        (l) Does $F4$ logically imply $F3$?






      Logically implies really confuses me. Attempt:



      Writing it in English first:



      F1: We have that $x$ never satisfies $P$ or there is a $y$ that satisfies $Q$



      F2: For some $x$ $P$ isn't satisfied, or $y$ always satisfies $Q$



      F3: Some $x$ doesn't satisfy $P$ or some $y$ satisfies $Q$



      F4: $x$ never satisfies $P$ or $y$ always satisfies Q



      By definition of logically implies: A formula $F$ logically implies a formula $F'$ iff every interpretation that satisfies $F$ satisfies $F'$



      So:



      (a) So for $F1$ $x$ never satisfies $P$ meaning it'll always be true whereas some $x$ can satisfy P in $F2$, thus this doesn't logically imply.



      (b) ... yeah I don't know how to explain any of this - any assistance even tips are appreciated I'm stumped



      Apparently $b$ is true I don't get it though.



      according to $F1$, $x$ never satisfies $P$ and by prepositional logic of $p to q$ is not p or q.



      then $F1$ is always true whereas $F3$ there can exist a P that satisfies Q and some y that doesn't satisfy Q so every interpretation doesn't satisfy? I'm understanding this poorly.







      predicate-logic






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      asked Dec 6 '18 at 17:31









      Tree GarenTree Garen

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