# Assignment 5

To appear....

### Exercise 6, pp. 94-5

6.1

Notes on notation: In the following answers, I will also use the symbol ``|'' for (left and right) angle brackets (to indicate ordered n-tuples). A denotes the universal quantifier, and E denotes the existential quantifier. Negation is symbolized by ``~''.

i. VM2,g(G(x1, x2, x3)) = 1

Since x1, x2, and x3 are free variables, in order to evaluate this formula, we need to check whether the ordered triple |g2(x1), g2(x2), g2(x3)| is an element of I2(G). According to the model, |g2(x1), g2(x2), g2(x3)| = |1, 5, 6|, which is an element of I2(G) (the set of triples |x,y,z| such that x + y = z), so the formula is in fact true in the model.

ii. VM2,g(Ax1G(x1, j, x1)) = 1

According to the truth conditions for the universal quantifier, the formula in (ii) is true just in case VM2, g[x/e](G(x1, j, x1)) = 1 for every e in the domain. In other words, Ax1G(x1, j, x1) is true just in case every way of assigning a value to x1 makes G(x1, j, x1) true. This clearly holds in the model M2, since the interpretation of G (I2(G)) is the set of triples such that the first plus the second equals the third, and the interpretation of j (I2(j)) is 0. The claim made by the formula in (ii), then, is every number plus 0 equals itself, which is obviously true.

iii. VM2,g(Ax2Ex3 [G(x1, x2, x3) v x2 = m]) = 1

What is important to notice about this example is that the variable x1 is free, while x2 and x3 are bound by the universal and existential quantifiers, respectively. This means that we need to use the assignment function g2 in order to get the value of x1, though the values of the other variables are determined by the semantics of the two quantifiers. The model tells us that the value of g2(x1) = 1 and the interpreation of the individual constant m is 9, therefore what the formula in (iii) says in ``logico-English'' is: for every number there is a second number such that adding 1 to the first gets the second, OR the first equals 9. This claim is true of everything in the domain U2 = {0,1,2,3, 4,5,6,7,8,9}, since the only value of x2 for which we can't find a value of x3 that makes the first part of the formula (G(x1, x2, x3)) true is 9, but this value of x2 makes the second part of the formula (x2 = m) true.

6.2

NOTE: In all of these exercises, expressions of the form ~x = y are supposed to mean ``x is not equal to y''. Of course, the fact that negation is a propositional connective (it only attaches to propositions) should tell you this, but in order to avoid confusion, it would have been better to have formalized the intended meaning with an expression like ~(x = y). The publisher and author have been notified....

iv. Ax1Ex2[K(x1, x2) & ~(x1 = x2)]
Every number is less than some number
This formula is false in M2, because 9 is not less than something.

Technically, this formula should be translated as ``everthing is less than something'', since there is no predicate in the logical form that corresponds to ``is a number''. However, since the domain contains only numbers, I will use expressions like every number and some number to make the English sentences sound more natural.

v. Ex1[Q(x1) & ~P(x1)] & Ex1[Q(x1) & P(x1)]
Bill Clinton uses an even number and an odd number in his lottery form.
This formula is true in M2 because the set of numbers that Bill Clinton uses in his lottery form contains both even and odd elements: I2(Q) = {0,1,2,3,4,5,7,8}.

vi. Ax1[[P(x1) & ~(x1 = m)] --> Q(x1)]
Every odd number other than 9 is in Bill Clinton's lottery form.
This sentence is indeed true in M2, since the numbers in the lottery form are {0,1,2,3,4,5,7,8}.

vii. Ax1[Q(x1) --> [P(x1) & ~(x1 = m)]]
Every number in Clinton's lottery form is an odd number other than 9.
This formula, however, is false, since Clinton's lottery form contains even numbers (as well as 0).

viii. Ax1[~Q(x1) --> [(x1 = m) v Ex2G(x2, x2, x1)]]
Every number that's not in Clinton's lottery form is either 9 or the sum of some other number plus itself.
The only two numbers that Clinton doesn't use in his lottery form are 9 and 6; since 6 is the sum of some other number plus itself (namely 3), this formula is true.

ix. Ex1[G(x1, x1, x1) & Ax2[G(x2, x2, x2) IFF (x2 = x1)]] (IFF denotes the biconditional.)
There is a unique number that equals the sum of itself plus itself.
This formula is true in the model: only 0 equals the sum of itself plus itself.

x. Ax1Ax2Ax3 [G(x1, x2, x3) --> [K(x2, x3) & ~(x2 = x3)]]
For every three numbers such that the first plus the second equals the third, the second is less than the third.
This formula is false: in the case in which the first number (x1) is 0, the values of the second and third numbers (x2 and x3) may be the same.

6.3

xi. Everything is odd or not odd.
Ax1[P(x1) v ~P(x1)]

xii. For every n, the thing that yeilds n when added to n is 0.
Ax1Ex2[G(x1,x2,x1) --> [(x2 = j) & Ax3[G(x1,x3,x1) --> (x3 = x2)]]]

At first glance, one might think that the English sentence (xii) should have a logical representation along the lines of (ii) above. This isn't so, however, because (xii) makes a stronger claim than (ii). (ii) says only that every number plus 0 equals itself; this allows for the possibility that there are other numbers that could be substituted in place of 0 and result in a true statement. The use of the definite article in (xii) eliminates this possibility, because it imposes a uniqueness condition. This is captured by the logical representation above, which says that for every number, if there is a second number such that the first plus the second equals the first, then the second is zero, AND any other number that derives the same result is equal to this one.

xiii. Everything is greater than or equal to itself.
Ax1[~L(x1, x1) v (x1 = x1)]

Since the negation of ``less than or equal to'' is ``greater than'', it is necessary to add the extra condition ``or (x1 = x1)'' in order to accurately characterize the truth conditions of the English sentence.

xiv. Every number Bill Clinton uses in his lottery forms is smaller than 9.
Ax1[Q(x1) --> [K(x1, m) & ~(x1 = m)]]

xv. Bill Clinton uses no more than four odd numbers in his lottery form.
Ax1[Q(x1) --> [Ex5Ex6 Ex7Ex8 [(Q(x5) & P(x5)) & (Q(x6) & P(x6)) & (Q(x7) & P(x7)) & (Q(x8) & P(x8))] --> ~P(x1)]]

Roughly speaking, this formula says: for every number that Bill Clinton uses in his lottery form, if you can find four numbers that are both odd and in the lottery form, then the first one is not odd. In other words, Bill Clinton uses no more than four odd numbers in his lottery form.

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