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How to prove that these propositions show that $f$ is injective?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How many non-equivalent formulas that use propositions p1… pn are there?Find two nonequivalent propositions which depend only on qWhat exactly do truth tables mean?Is the Cardinality of the Set of Contingent Propositions the Same as the Cardinality of the Set of Tautologies?Is an argument valid simply if its form is valid?Check if a proposition with a equation is true or false and then write its negation.Predicates and QuantifiersWhich of these are propositions? Comment on their truth values if not, give a reason why it is not.Is “$x$ is an integer, $1 < x < 3$, and $x^2 = 4$” a proposition?logical equivalence between statements










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Let $f$ be a total function on some nonempty set $D$. In the following propositions, x and y are variables ranging over $D$, and g is a variable ranging over total functions on $D$. Indicate all of the propositions that are equivalent to the proposition that $f$ is an injection.



(1) $exists g,forall x : (g circ f) (x) = x $



(2) $exists g,forall x : (f circ g) (x) = x $



I know that (1) is true and (2) is false, but I don't understand why. Could someone please explain?
Thank you.










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    Let $f$ be a total function on some nonempty set $D$. In the following propositions, x and y are variables ranging over $D$, and g is a variable ranging over total functions on $D$. Indicate all of the propositions that are equivalent to the proposition that $f$ is an injection.



    (1) $exists g,forall x : (g circ f) (x) = x $



    (2) $exists g,forall x : (f circ g) (x) = x $



    I know that (1) is true and (2) is false, but I don't understand why. Could someone please explain?
    Thank you.










    share|cite|improve this question









    $endgroup$














      1












      1








      1





      $begingroup$


      Let $f$ be a total function on some nonempty set $D$. In the following propositions, x and y are variables ranging over $D$, and g is a variable ranging over total functions on $D$. Indicate all of the propositions that are equivalent to the proposition that $f$ is an injection.



      (1) $exists g,forall x : (g circ f) (x) = x $



      (2) $exists g,forall x : (f circ g) (x) = x $



      I know that (1) is true and (2) is false, but I don't understand why. Could someone please explain?
      Thank you.










      share|cite|improve this question









      $endgroup$




      Let $f$ be a total function on some nonempty set $D$. In the following propositions, x and y are variables ranging over $D$, and g is a variable ranging over total functions on $D$. Indicate all of the propositions that are equivalent to the proposition that $f$ is an injection.



      (1) $exists g,forall x : (g circ f) (x) = x $



      (2) $exists g,forall x : (f circ g) (x) = x $



      I know that (1) is true and (2) is false, but I don't understand why. Could someone please explain?
      Thank you.







      propositional-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 25 at 19:54









      M. RoshidM. Roshid

      323




      323




















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Hint 1



          Assume that $f(x)=f(y)$ with $xne y.$ Then



          $$x=g(f(x))=g(f(y))=y$$ gives a contradiction.






          share|cite|improve this answer









          $endgroup$













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            1












            $begingroup$

            Hint 1



            Assume that $f(x)=f(y)$ with $xne y.$ Then



            $$x=g(f(x))=g(f(y))=y$$ gives a contradiction.






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              Hint 1



              Assume that $f(x)=f(y)$ with $xne y.$ Then



              $$x=g(f(x))=g(f(y))=y$$ gives a contradiction.






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                Hint 1



                Assume that $f(x)=f(y)$ with $xne y.$ Then



                $$x=g(f(x))=g(f(y))=y$$ gives a contradiction.






                share|cite|improve this answer









                $endgroup$



                Hint 1



                Assume that $f(x)=f(y)$ with $xne y.$ Then



                $$x=g(f(x))=g(f(y))=y$$ gives a contradiction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 25 at 20:22









                mflmfl

                26.9k12142




                26.9k12142



























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