Logic Question 1:
Show that (V)(32)Lwz' and '(z)(w)Lwz' are not quantificationally equivalent by showing that the interpretation, 1 (specified below), makes one of the sentences true and not the other. A demonstration is largely completed below, so all you will need to do is fill in the ellipses. (So just turn in the wording you put in the ellipses.) Proof Let I be such that: UD: { Noam, Robert, Willard, Nelson } Lxy: 'Lxy'is the set of pairs (x,y) such that x likes y. On I, Noam and Robert like each other, but neither likes Willard or Nelson. And Willard and Nelson like each other, but neither likes Noam or Robert. (And each likes himself if you were wondering.) By our definition of truth on an interpretation, '(VW)z)Lwz' is true on I iff for every assignment of values to the variables, for every way of shifting the value of w' to some member, u, of UD, there is some way of shifting the value of 'z' to a member, u', of UD such that (u, u') is in the extension of L". Indeed, (no matter the variable assignment) if the value of 'w' is shifted to any of Noam, Robert, Willard, or Nelson, then there is some way of shifting the value ofz to an object in UD that Noam, or Robert, or Willard, or Nelson likes. In other words, on I, everybody likes someone or other. So I makes (Vw)(Iz)Lwz' true. On the other hand, "(Ez) Vw)Lwz' is false on I. This is because the definition of truth on an interpretation tells us that I makes "(Iz)(Vw)Lwz' true iff[...]. But no matter whether the value of *2' is shifted to Noam, Robert, Willard, or Nelson, there is some shift of the value of 'w' to some object u such that [...]. In other words, on I it is not the case that there exists a thing that everybody likes. a Show that (V)(32)Lwz' and '(z)(w)Lwz' are not quantificationally equivalent by showing that the interpretation, 1 (specified below), makes one of the sentences true and not the other. A demonstration is largely completed below, so all you will need to do is fill in the ellipses. (So just turn in the wording you put in the ellipses.) Proof Let I be such that: UD: { Noam, Robert, Willard, Nelson } Lxy: 'Lxy'is the set of pairs (x,y) such that x likes y. On I, Noam and Robert like each other, but neither likes Willard or Nelson. And Willard and Nelson like each other, but neither likes Noam or Robert. (And each likes himself if you were wondering.) By our definition of truth on an interpretation, '(VW)z)Lwz' is true on I iff for every assignment of values to the variables, for every way of shifting the value of w' to some member, u, of UD, there is some way of shifting the value of 'z' to a member, u', of UD such that (u, u') is in the extension of L". Indeed, (no matter the variable assignment) if the value of 'w' is shifted to any of Noam, Robert, Willard, or Nelson, then there is some way of shifting the value ofz to an object in UD that Noam, or Robert, or Willard, or Nelson likes. In other words, on I, everybody likes someone or other. So I makes (Vw)(Iz)Lwz' true. On the other hand, "(Ez) Vw)Lwz' is false on I. This is because the definition of truth on an interpretation tells us that I makes "(Iz)(Vw)Lwz' true iff[...]. But no matter whether the value of *2' is shifted to Noam, Robert, Willard, or Nelson, there is some shift of the value of 'w' to some object u such that [...]. In other words, on I it is not the case that there exists a thing that everybody likes. a
