*Introduction to Logic*

*x*in the domain of

*f*

*f*

^{–1}(

*f*(

*x*)) =

*x*,

*x*in the range of

*f*

*f(f*

^{–1}(

*x*)) =

*x*.

*f*

^{–1}(

*x*)’ for ‘

*x*’; (ii) Apply (II);(iii) Solve the resulting equation for ‘

*f*

^{–1}(

*x*)’.

Besides these operations, which are all simply special cases of the operations on sets, it is also possible to define some special operations on binary relations which depend on the fact that they are sets of ordered couples. The converse of a relationR

is the relation such that, for allxandy,xRyif and only ifyRx. Thus the converse of a relation is obtained simply by reversing the order of all the ordered couples which constitute it. The converse of the relationHwhich holds betweenxandyif and only ifxis the husband ofy, is the relationWwhich holds betweenyandxif and only ifyis the wife ofx. Thus,

Since a function is simply a special kind of relation, we can speak of the converse of a function. The converse of a function is always a relation, but in general it will not be a function. |(Suppes 234-235)

*f*(

*x*)=

*x*

^{2}. Now suppose we will render this as a relation whose set is ordered

*n*-tuples. My first thought would be that the ordered tuples for <

*x*,

*y*> would be things like <2, 4>, <3, 9> and so on, because I would think that it would take a number from the domain and assign to it the square of that number. But he says that in this case

*x*=

*y*

^{2}. That is probably correct, but I still do not know why, because as I understand it,

*y*=

*x*

^{2}, when we take

*x*to be the first number of the couple and

*y*to be the second number which is the square of the first. At any rate, if we keep that structure where the first number comes first in the couple and the second is its square, then the rest of his example makes sense to me.]

A a second example, supposefis the function such that for every real numberx,f(x)=x^{2}if we considerfas a relation, <x,y>∈fif and only ifx=y^{2}.Consider, now,x= 4. Thenand also <4, –2>∈f. Henceis not a function.(Suppes 235)

*inverse*of the first one, and we use the superscript ‘–1’ to notate it instead of the cup symbol.

On the other hand, some functions are such that their converses are also functions. For example, ifgis the function such that for every real numberx,g(x) = 2x+ 4.When a functionfis such that its converse relationis also a function, then we say thatis theinverseoff. To mark this special situation we use ‘–1’ as a superscript in place of the more general cup notationThus if the converse offis a function,We use the notation^{‘–1’ }only whenis a function.(Suppes 235)

g(x) = 2x+ 4.(Suppes 235)

*g.*So

*g*(2) = 12. Now, let us take that 12, and apply it to

*g*

^{–1}. So

*g*

^{–1}(12) = 2, which is what we started with. Or another way to write that could be:

*g*

^{–1}(

*g*(2)) = 2.]

It should be noticed that when the inverse offexists (i.e., the converse offis a function), we have the following identities: For everyxin the domain off(I)f^{–1}(f(x)) =x,and for everyxin the range off(II)Principles (I) and (II) are both important in solving forf(f^{–1}(x)) =x.fandf^{–1}as we shall soon see.

(Suppes 235)

*one-one*” (235). [I suppose the idea is this. Think of a relation that relates each term in the domain with its own assigned one in the counterdomain. Possibly it is many-one, meaning that more than one item in the domain is assigned the same one in the counterdomain. Suppose also that it is one-many. That means, to each member of the counterdomain is assigned a unique member of the domain. This I think normally means that more than one member of the counterdomain could get various members of the domain. Howeve, since we are combining them, for any term in the domain, there will only be one unique member in the counterdomain it is related to, and likewise, for any in the counterdomain, there is only one in the domain, and furthermore, that relation is the same going both ways. So it is one-one now.]

At the end of the last section we spoke of many-one and one-many relations. A relation which is both many-one and one-many isone-one.(Suppes 235)

*R*, and afterward he will determine which are equivalent.]

We also speak of a one-one relation as a one-one function. Consider now the following statements about a relationR:

(1)Ris a function.(2)Ris a many-one relation.(3) The converse ofRis a function.(4)Ris a one-many relation.(5)Rand its converse are both functions.(6)Ris a function and its inverse exists.(7)Ris a many-one relation and R is a one-many relation.(8)Ris a one-one relation.(9)(Suppes 236)Ris a one-one function.

*R*is a function’ ↔ ‘

*R*is a many-one relation’. We saw this already in the definition of function. A function is one that assigns to a member of the domain a unique member of the counterdomain. 3 and 4 are equivalent, so ‘The converse of

*R*is a function’ ↔ ‘

*R*is a one-many relation’. If the converse is a function, that means to each of the original function’s counterdomain members is assigned a unique member of the domain, and thus the original

*R*relation is a one-many relation. Finally, 5 through 9 are all equivalent. As we see from them, the reasoning for this is found in the definitions and discussions we just had.]

The following equivalences hold:

(1) ↔ (2),

(3) ↔ (4),

(5) ↔ (6) ↔ (7) ↔ (8) ↔ (9).

(Suppes 236)

(1)f(x) = 5x– 4

*x*is a real number. There are three phases to this procedure. In the first phase, we substitute

*f*

^{–1}(

*x*) for

*x*. In our example, he notes that

Since (1) holds for every real numberx, we may substitute‘f^{–1}(x)’ for ‘x’ (application of universal specification) and obtain:(2)f(f^{–1}(x)) = 5f^{–1}(x) – 4.

Rule of Universal Specification:US.If a formulaSresults from a formulaRby substituting a termtfor every free occurrence of a variablevinRthenSis derivable from(v)R.In the above derivation, the term t is simply ‘(Suppes 59-60)x’ in both cases of universal specification, which is why we spoke ofdroppingquantifiers, but in general | we want to be able to replace the quantified variable by any term. Thus if ‘a’ is the name of John Quincy Adams, we may infer ‘Aa→Ma’, as well as ‘Ay→My’, from ‘(x)(Ax→Mx)’. And in the case of arithmetic from the statement ‘(x)(x+ 0 =x)’ we may infer by universal specification ‘(x+ y) + 0 =x+ y’; in this case the term t is ‘x+ y’.

*‘f*

^{–1}(

*x*)’ for cases of ‘

*x*’

(1)f(x) = 5x– 4(2)f(f^{–1}(x)) = 5f^{–1}(x) – 4.

*x*in the range of

*f*

(II)f(f^{–1}(x)) =x.

*f(f*

^{–1}(

*x*)) we then place

*x*.

(2)f(f^{–1}(x)) = 5f^{–1}(x) – 4(3)x= 5f^{–1}(x) – 4(Suppes 236)

*f*(

^{–1}*x*)). [So let us place the formula we need to solve for:

x= 5f^{–1}(x) – 4

*f*

^{–1}(

*x*) from both sides.

x–5f^{–1}(x) = – 4

*x*from both sides.

–5f^{–1}(x) =–x – 4

5f^{–1}(x) =x + 4

*f*

^{–1}(

*x*).]

(Suppes 236)

(5)f(x) =x^{2 }– 1(Suppes 237)

*f*

^{–1}(

*x*) for

*x*.

f(f^{–1}(x)) =f^{–1}(x)^{2 }– 1

(II)f(f^{–1}(x)) =x.

(6)x=f^{–1}(x)^{2 }– 1

*f*(

^{–1}*x*)). I might get this wrong, but let us start by subtracting

*f*

^{–1}(

*x*)

^{2 }from both sides.

^{ }–f^{–1}(x)^{2 }+x= – 1

*x*from both sides.

^{ }–f^{–1}(x)^{2 }= –x– 1

^{ }f^{–1}(x)^{2 }=x+ 1

(Suppes 237)

*x*, we get the square root of four, which is two and negative two.]

We thus infer that^{ }f^{–1}(3)^{ }= 2 and^{ }f^{–1}(3)^{ }= –2and hence2 = –2which is absurd.(Suppes 237)

*f*

^{–1}(

*x*) in the equation, it is not a term that we can substitute using universal specification, because that requires the term being substituted to be consistently itself throughout. I think I misunderstand this, so let me quote.]

The mistake here was in inferring (6) from (5). The substitution of ‘f^{–1}(x)’ for ‘x’ is permissible only if ‘f^{–1}(x)’ is a proper term. Iffis not a function, the expression ‘f^{–1}(x)’ is not a term, since it does not designate a unique entity.(Suppes 237)

*f*(

*x*)=

*x*

^{2}. This will “output” a series of values, depending on the “input” values. The mathematical inverse needs just to be one that outputs the original series of values. For

*f*(

*x*)=

*x*

^{2}this would be the square root of

*x*. The fact that this function now outputs negated terms along with their non-negated counterparts does not disqualify it. Suppes then discusses the solution they have in mathematics, where apparently they arbitrarily select one value as the ‘principle’ one. This part of the explanation is a too technical for me at the moment. Since this concerns mathematics more than it seems to concern logic, I will not complicate the summary by including it. See pages 237-238 for the details.]

*Relative product of two relations.*This operation combines two separate relations by finding ones where the second member of one couple (in the first relation) is the same as the first member of the second couple (in the other relation). From these pairings of couples, you make a new couple that takes the first member of the first couple and the second member of the second couple. Being an aunt is an example, with you and your aunt being the end-terms of separate relations (you being the child of your parent, and your parent being the sibling of your aunt), with your parent being the middle term that gets excluded.

IfRandSare binary relations, then by the relative product ofRandS(in symbols:R/S) we mean the relation which holds betweenxandyif and only if there exists azsuch thatRholds betweenxandz, andSholds betweenzandy. Symbolically,xR/S y↔ (∃z) (xRz&zSy).IfxPymeans thatxis a parent ofy, andxSymeans thatxis a sister ofy, thenx(S/P)ymeans that there is azsuch thatxis a sister ofzandzis a parent ofy, and hence such thatxis an aunt ofy.(Suppes 226)

*f*(

*g*(

*x*)) or it can be considered as one whole function like

*h*(

*x*). Suppes will use a notation system where a small circle is used between the two composed functions. Now, suppose we used the relative product notation of the slash to notate this. Suppose we have two functions: (a)

*f*(x) =

*x*+ 2 and

*g*(

*x*) =

*x*

^{2}. And suppose our composition is of this form:

*f*(

*g*(x)). And for the sake of comparison with relations, let us turn these into ordered couples. We will take

*x*here to be 2. So our first couple is <2, 4>. We then have <4, 6> from function

*f*. Now were we to use the slash formulation, we would keep the structure of <2, 4> / <4, 6> and write it

*g*/

*f.*Suppes says the problem with this is that it is not the right natural order for

*f*(

*g*(x)). So he uses a circle:

*f*∘

*g*. Suppes will then say that the composition of functions is associative but not commutative. That it is not commutative would seem to result from the fact that if you change the order of the functions you will get different results and so it is not overall the same function. I am not sure about the associative idea, but perhaps this is for when you have a composition of for example three functions, and were you to compute the value, you could either do the first pair first or the second pair second. However, I do not know how that would work, because I would think you always need to begin with the function that is most nested, as it gives the value for the next function, and so on. Let me quote:]

The relative product of two functions plays an important role in many discussions and is sometimes given a special name: thecompositionof the functions. Suppose thatfandgare functions, and thatx,y, andzare entities such that(7)y=f(z)and(8)z=g(x);thenxstands toyin the relationg/f. In other terms, from equations (7) and (8) we see, thaty=f(g(x)).Thus the relative product, or composition, of two functions is a functionhsuch that (wheneverxis in the domain of definition ofg, andg(x) is in the domain of definition off)h(x)f(g(x)).This operation of forming the composition of two functions is so extensively used in certain branches of mathematics that various special symbols have been used for it; we shall use a small circle ‘∘’. Thus, (f∘g)(x) =f(g(x)). We introduce this special notation instead of using the relative product notation because the order of ‘f’ and ‘g’ in ‘f∘g’ is the natural one corre- | sponding to their order in ‘f(g(x))’, whereas this order is reversed in the relative product:f∘g=g/f.It is clear that composition of functions is associative but not commutative.(Suppes 238-239)

The problem of finding the inverse of a function is actually a special case of the problem of finding “the” functiong, given the functionsfandf∘g. In finding the inverse,gisf^{–1}, and(f∘g)(x) = (f∘f^{–1})(x) =f(f^{–1}(x)) =x.(Suppes 239)

*g*is not unique. I do not follow these ideas so well. I think he means that

*g*does not assign to

*x*a unique element, but I am not sure. He then says that the problem is to “find at least one function

*g*such that

*f*∘

*g*is identical with some given function”. I think the idea is that we have function

*f*. We want to find its inverse, which we will call function

*g*. The next part confuses me, because we now deal with two separate functions, and I do not see how we are trying to find an inverse. Instead, the problem is set up as that we have two functions,

*f*and

*h*, and we want to know, what third function does

*f*need to operate on in order to get

*h*? Suppes will give two techniques for finding such functions. As I do not follow some steps, and also as it seems to be more of a technique for solving mathematical sorts of problems, I will leave this part out of the summary. But for details, see pages 239-240.]

*Introduction to Logic*. New York: Van Nostrand Reinhold / Litton Educational, 1957.

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