What is the difference in meaning between these two antecedents …?

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$$ (forall x)(Mx to Wx) to quad... tag1 $$
$$ (forall x)(Mx land Wx) to quad... tag2 $$



The consequent is clear:
$, (exists y)(Fy land Sy)$



The statement is:




If every member of the bar is wrong, then there is a federal court which will sustain the decision.





propositional-calculus first-order-logic predicate-logic formal-proofs logic-translation




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    up vote
    2
    down vote

    favorite












    $$ (forall x)(Mx to Wx) to quad... tag1 $$
    $$ (forall x)(Mx land Wx) to quad... tag2 $$



    The consequent is clear:
    $, (exists y)(Fy land Sy)$



    The statement is:




    If every member of the bar is wrong, then there is a federal court which will sustain the decision.





    propositional-calculus first-order-logic predicate-logic formal-proofs logic-translation




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      $$ (forall x)(Mx to Wx) to quad... tag1 $$
      $$ (forall x)(Mx land Wx) to quad... tag2 $$



      The consequent is clear:
      $, (exists y)(Fy land Sy)$



      The statement is:




      If every member of the bar is wrong, then there is a federal court which will sustain the decision.





      propositional-calculus first-order-logic predicate-logic formal-proofs logic-translation




      share|cite|improve this question











      $$ (forall x)(Mx to Wx) to quad... tag1 $$
      $$ (forall x)(Mx land Wx) to quad... tag2 $$



      The consequent is clear:
      $, (exists y)(Fy land Sy)$



      The statement is:




      If every member of the bar is wrong, then there is a federal court which will sustain the decision.






      propositional-calculus first-order-logic predicate-logic formal-proofs logic-translation





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      share|cite|improve this question




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      asked Jul 16 at 7:47









      Ali

      386




      386




















          2 Answers
          2






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          down vote



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          The difference is that :




          $(∀x)(Mx ∧ Wx)$




          means :




          "every (object in the universe) is a member of the bar and is wrong".




          This is not what we want asserting the conditional with antecedent :




          "If every member of the bar is wrong, ..."




          The first one is FALSE if there is somewhere an individual that is not a member of the bar, while the second one is TRUE also if there are individuals that are not members of the bar, provided that every individual that is member of he bar is wrong.






          share|cite|improve this answer






























            up vote
            0
            down vote













            The first one is what you want. It says that if $x$ is a member of the bar, then $x$ is wrong, which is what's intended. The second one says everyone is a member of the bar, and wrong.






            share|cite|improve this answer





















            • but the symbol $(to)$ is there
              – Ali
              Jul 16 at 7:56










            • @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
              – saulspatz
              Jul 16 at 7:59











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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










            The difference is that :




            $(∀x)(Mx ∧ Wx)$




            means :




            "every (object in the universe) is a member of the bar and is wrong".




            This is not what we want asserting the conditional with antecedent :




            "If every member of the bar is wrong, ..."




            The first one is FALSE if there is somewhere an individual that is not a member of the bar, while the second one is TRUE also if there are individuals that are not members of the bar, provided that every individual that is member of he bar is wrong.






            share|cite|improve this answer



























              up vote
              2
              down vote



              accepted










              The difference is that :




              $(∀x)(Mx ∧ Wx)$




              means :




              "every (object in the universe) is a member of the bar and is wrong".




              This is not what we want asserting the conditional with antecedent :




              "If every member of the bar is wrong, ..."




              The first one is FALSE if there is somewhere an individual that is not a member of the bar, while the second one is TRUE also if there are individuals that are not members of the bar, provided that every individual that is member of he bar is wrong.






              share|cite|improve this answer

























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                The difference is that :




                $(∀x)(Mx ∧ Wx)$




                means :




                "every (object in the universe) is a member of the bar and is wrong".




                This is not what we want asserting the conditional with antecedent :




                "If every member of the bar is wrong, ..."




                The first one is FALSE if there is somewhere an individual that is not a member of the bar, while the second one is TRUE also if there are individuals that are not members of the bar, provided that every individual that is member of he bar is wrong.






                share|cite|improve this answer















                The difference is that :




                $(∀x)(Mx ∧ Wx)$




                means :




                "every (object in the universe) is a member of the bar and is wrong".




                This is not what we want asserting the conditional with antecedent :




                "If every member of the bar is wrong, ..."




                The first one is FALSE if there is somewhere an individual that is not a member of the bar, while the second one is TRUE also if there are individuals that are not members of the bar, provided that every individual that is member of he bar is wrong.







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited Jul 16 at 8:09


























                answered Jul 16 at 8:00









                Mauro ALLEGRANZA

                60.7k346105




                60.7k346105




















                    up vote
                    0
                    down vote













                    The first one is what you want. It says that if $x$ is a member of the bar, then $x$ is wrong, which is what's intended. The second one says everyone is a member of the bar, and wrong.






                    share|cite|improve this answer





















                    • but the symbol $(to)$ is there
                      – Ali
                      Jul 16 at 7:56










                    • @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
                      – saulspatz
                      Jul 16 at 7:59















                    up vote
                    0
                    down vote













                    The first one is what you want. It says that if $x$ is a member of the bar, then $x$ is wrong, which is what's intended. The second one says everyone is a member of the bar, and wrong.






                    share|cite|improve this answer





















                    • but the symbol $(to)$ is there
                      – Ali
                      Jul 16 at 7:56










                    • @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
                      – saulspatz
                      Jul 16 at 7:59













                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    The first one is what you want. It says that if $x$ is a member of the bar, then $x$ is wrong, which is what's intended. The second one says everyone is a member of the bar, and wrong.






                    share|cite|improve this answer













                    The first one is what you want. It says that if $x$ is a member of the bar, then $x$ is wrong, which is what's intended. The second one says everyone is a member of the bar, and wrong.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered Jul 16 at 7:52









                    saulspatz

                    10.7k21323




                    10.7k21323











                    • but the symbol $(to)$ is there
                      – Ali
                      Jul 16 at 7:56










                    • @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
                      – saulspatz
                      Jul 16 at 7:59

















                    • but the symbol $(to)$ is there
                      – Ali
                      Jul 16 at 7:56










                    • @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
                      – saulspatz
                      Jul 16 at 7:59
















                    but the symbol $(to)$ is there
                    – Ali
                    Jul 16 at 7:56




                    but the symbol $(to)$ is there
                    – Ali
                    Jul 16 at 7:56












                    @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
                    – saulspatz
                    Jul 16 at 7:59





                    @Ali Yes, and that symbol means "if...then." For every $x,$ if $Mx$ then $Wx$ that is, "for every $x,$ if $x$ is a member of the bar, then $x$ is wrong." Surely this is the same as saying "every member of the bar is wrong."
                    – saulspatz
                    Jul 16 at 7:59













                     

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