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[Griss, “Negationless Intuitionistic Mathematics, I”, entry directory]
[The following is summary. I am not a mathematician, so please consult the original text instead of trusting my summarizations, which are surely mistaken or inelegantly articulated. Bracketed comments and subsection divisions are my own. Proofreading is incomplete, so please forgive my mistakes.]
Summary of
George François Cornelis Griss
(G.F.C. Griss)
“Negationless Intuitionistic Mathematics, I”
1.2
“Properties of the Relations ‘The Same’ and ‘Different’”
Brief summary:
(1.2.1) The first property of sameness and difference for our intuitionally and non-negationally constructed sets of natural numbers is that: Two elements of the set {1, 2, ..., m} are the same or distinguishable. (1.2.2) The second property of sameness and difference is that if two numbers (which may either be the same or different numbers, but we do not determine that initially) share all the same differences to all the other numbers, then they are the same number (or if they are unequal to all the other same numbers, then they are equal to one another): “If for two elements a and b of {1, 2 ..., m} holds: a ≠ c for each c ≠ b, then a = b.” (1.2.3) The complementary set of the element a of the set {1, 2, ..., m} is denoted by A. And “The complement of A is a and the sum of a and A is {1, 2, ..., m}”. The “main proposition of arithmetic” would be formulated here as: “If there is a one to one reciprocal correspondence between {1, 2, ..., m} and {1, 2, ..., p}, then m = p.” “For the elements of the set {1, 2, ..., m} the following propositions hold now:
I a = a
II a = b → b = a
III a = b and b = c → a = c
IV a ≠ b → b ≠ a
V a = b and b ≠ c → a ≠ c
VI a = b or a ≠ b
VII a ≠ c for each c ≠ b → a = b.
Proposition “VI replaces the negative proposition: Two natural numbers are the same or not,” which holds in non-intuitionistic mathematics but not in intuitionistic mathematics, on account of the principle of excluded middle or excluded third not holding. Proposition VII is functionally correspondent with its negational counterpart, which is: “If it is impossible, that a is not the same as b, then a is the same as b.” And our positive theory replaces the following other negational propositions regarding sameness and difference:
different ⇄ not the same.
the same ⇄ not different.
the same and different exclude one another.
two natural numbers are either the same or different.
[The Being Either Same or Different for Natural Numbers]
[Equality as Shared Difference]
[Remaining Properties]
Summary
[The Being Either Same or Different for Natural Numbers]
[The first property of sameness and difference for our intuitionally and non-negationally constructed sets of natural numbers is that: Two elements of the set {1, 2, ..., m} are the same or distinguishable.]
[In the previous section 1.1, we constructed sets of natural numbers {1, 2, 3, ..., n} inductively by beginning with a selfsame object, 1, which is the same as itself, so 1 = 1. Then we added another number, 2, which is also selfsame, so 2 = 2, but it is distinguishable from 1, so 1 ≠ 2, 2 ≠ 1. When we added 3, which is also selfsame and also distinct from the other two, so 3 = 3 and 1 ≠ 3, 3 ≠ 1, 2 ≠ 3, 3 ≠ 2. Now, we think of some object whose identity we do not assume beforehand but which we will deduce on the basis of its samenesses and differences. We consider a set of any number of members, {1, 2, …, n} and more broadly a set that is one member larger, {1, 2, …, n′}. And we say that if some given number whose identity we do not begin by assuming is distinguishable from each element of {1, 2, …, n}, then it is n′, but if it is distinguishable from n′, then it is a member of {1, 2, …, n}. Also, if we cease our additions with an mth element, we get a finite set {1, 2, …, m}. Now Griss will outline some of the properties of these same and different relations. (They seem closely tied to equality and inequality of quantity, but I am not sure how to distinguish sameness and equality, and difference and inequality. The other idea at work here is distinguishability. But that still does not help me, because how would we distinguish sameness, equality, and indistinguishability from one another, and how would we distinguish difference, inequality, and distinguishability from one another?) The first property of sameness and difference is that any two members of a finite set of natural numbers is either the same or distinguishable. (I am not sure how any two can be the same, if we have fashioned the set on the basis of adding distinct terms consecutively. Perhaps the idea is that we can have both 2 and 2, or maybe, we have two variables, a and b, whose identity is not initially assumed but whose sameness is deduced.) (The proof for this I may not get right, but it seems to work in the following way.) We will show that regardless of what might be in the set, in any case any two elements a and b will be either the same or different. We begin by saying that the proposition holds for {1, 2}, (but no explanation is given for that. Perhaps the idea is that with only two members, there can only be two possibilities for a and b, either they are both 1 or both 2, or one is 1 and the other is 2.) The proof proceeds to the larger set {1, 2, ..., n} and it seems also to a one-larger set {1, 2, ..., n′}. So with this larger set, we denote two elements as a and b. Since they are both in the larger set {1, 2, ..., n′}, that means they can be either in the subset {1, 2, ..., n} or be the only remaining item n′. For our convenience of illustration, we will have the set {1, 2, 3} with 2 exemplifying n, and 3 being n′. There are four possible situations that may hold. [1] both a and b equal n′ (they both are 3). Thus the proposition holds, because the two elements are the same. [2] a is n′ (a is 3), and b belongs to the set of other numbers (b is either 1 or 2). In this case, the proposition holds, because the two elements are different. [3] a belongs to the set of other numbers (a is either 1 or 2), and b is n′. Again, the proposition holds, because here a and b are different. And [4], a and b both belong to the set of other numbers (they are either 1 or 2), in which case either a is the same as b (they are both either 1 or both 2), or they are different (a is one of the two options, and b is the other option.) Here we again see that the proposition holds, because in this case a and b are either the same or different. As these are the only possibilities, it is necessarily the case that two elements of such a set are either the same or distinguishable.]
Proposition: Two elements of the set {1, 2, ..., m} are the same or distinguishable.
Proof: For {1, 2} the proposition holds. Let the proof have proceeded to the set {1, 2, ..., n}. Denote two elements of {1, 2, ..., n′} by a and b. a is an element of {1, 2, ..., n} or it is n′, likewise b. There are 4 possibilities: 1) a = n′ and b = n′, so a = b; 2) a = n′ and b belongs to {1, 2, ..., n}, so a ≠ b; 3) a belongs to {1, 2, ..., n} and b = n′, so a ≠ b; 4) a and b belong to {1, 2, ..., n}, then a = b or a ≠ b.
(1131)
[Equality as Shared Difference]
[The second property of sameness and difference is that if two numbers (which may either be the same or different numbers, but we do not determine that initially) share all the same differences to all the other numbers, then they are the same number (or if they are unequal to all the other same numbers, then they are equal to one another): “If for two elements a and b of {1, 2 ..., m} holds: a ≠ c for each c ≠ b, then a = b.”]
[The next proposition regarding the properties of sameness and difference is something like the following. If two unidentified elements of a set of natural numbers share all the same differences to the other numbers (if they are different to all the other same numbers), then they are the same number. So suppose our set has three numbers, {1, 2, 3}. And we have two unidentified numbers in that set, a and b, which are either the same number or are two different numbers. We will show the conditions under which they are the same, non-negationally. For this, we need to consider a third number c, and ask about c’s relation to a and b. For the sake of illustration, let us just assume that both a and b are the number 2, so that we can see how this proposition works. So b is 2. We now want to see all the other terms that b does not equal, that is to say, all the c’s that it is distinct from. In this case, they are 1 and 3. The proposition says, “If for two elements a and b of {1, 2 ..., m} holds: a ≠ c for each c ≠ b, then a = b.” So we have determined each c ≠ b, with those c’s being 1 and 3. Now, for each such inequality, namely, 1 ≠ b and 3 ≠ b, where 1 and 3 are the c’s, we also have that a as well does not equal those same c’s: 1 ≠ a and 3 ≠ a. So since a and b are different from the same set of all the other members, they must be identical to each other. For, there is only one option left that both can be, which is 2. Now let us look at the actual proof. We begin by asserting that the proof holds for the set {1, 2}, but I again am not sure the reasoning why. Perhaps it is because c can never be a third number. So if both a and b are different than c, then they must be the number other than c, and with there only being one option, both a and b would have to be that same one. But I am not sure. We then proceed to a larger set, {1, 2, ..., n} and it seems more so to {1, 2, ..., n′}. We can say that b either belongs to {1, 2, ..., n} or b = n′. So we can now say that there are then two possibilities with regard to the difference relations between a and b with respect to c. [1] One possibility is that b = n′. (From the text below, which you should consult, I do not grasp all the reasoning, but I am guessing it is the following. We are assuming that a ≠ c for each c ≠ b. And we are saying here that b = n′, which means that b cannot be in the set {1, 2, ..., n}. From this we are supposed to conclude that a is distinguishable from each element of {1, 2, ..., n}. But I cannot see how that can be inferred from b = n′. This is why I wonder if we need to include the other assumptions. So perhaps {1, 2, ..., n} gives us all the c’s that do not equal b. And we are also assuming that a does not equal any of these c’s. That means a cannot be in that set, leaving the only other option, it being n′, which b is also, and thus a = b.) [2] The second possibility is that b does in fact belong in the set {1, 2, ..., n}. We next want to know about the numbers c that b does does not equal. We know at least one, namely n′, because it lies outside the set that b is a member of. So we can say that one such c would have to be n′. Now, one of our assumptions was that a ≠ c for each c ≠ b. So we can say that a is also not n′ and that a is a member of {1, 2, ..., n} just like b is. But from this, we are to conclude that a = b, which is not yet obvious to me. Could not a = 1 and b = 2? Both are in that set. So to come to that conclusion, I wonder if you repeat the exercise. After establishing that both a and b are not n′, then perhaps we must then establish all the other numbers in the set {1, 2, ..., n} that b is not, going one by one, each time noting that a is not equal to that number, until finally we arrive upon the one same number that they both must be. I am not sure actually. So please consult the quotation below.)]
Proposition: If for two elements a and b of {1, 2 ..., m} holds: a ≠ c for each c ≠ b, then a = b.
Proof: For {1, 2} the proposition holds. Let the proof have proceeded to {1, 2, ..., n}. There are two possibilities: In {1, 2, ..., n′} b = n′ or b belongs to {1, 2, ..., n}. 1) If b = n′, then a is distinguishable from each element of {1, 2, ..., n}, so a = n′ and a = b. 2) b belongs to {1, 2, ..., n}; take c = n′, then a also belongs to {1, 2, …, n}, so a = b.
(1132, boldface is mine)
[Remaining Properties]
[The complementary set of the element a of the set {1, 2, ..., m} is denoted by A. And “The complement of A is a and the sum of a and A is {1, 2, ..., m}”. The “main proposition of arithmetic” would be formulated here as: “If there is a one to one reciprocal correspondence between {1, 2, ..., m} and {1, 2, ..., p}, then m = p.” “For the elements of the set {1, 2, ..., m} the following propositions hold now:
I a = a
II a = b → b = a
III a = b and b = c → a = c
IV a ≠ b → b ≠ a
V a = b and b ≠ c → a ≠ c
VI a = b or a ≠ b
VII a ≠ c for each c ≠ b → a = b.
Proposition “VI replaces the negative proposition: Two natural numbers are the same or not,” which holds in non-intuitionistic mathematics but not in intuitionistic mathematics, on account of the principle of excluded middle or excluded third not holding. Proposition VII is functionally correspondent with its negational counterpart, which is: “If it is impossible, that a is not the same as b, then a is the same as b.” And our positive theory replaces the following other negational propositions regarding sameness and difference:
different ⇄ not the same.
the same ⇄ not different.
the same and different exclude one another.
two natural numbers are either the same or different.]
[The complementary set of an element plus that element is their sum, making the full set. “If we denote the complementary set of the element a of the set {1, 2, ..., m} by A, the complement of A is a and the sum of a and A is {1, 2, ..., m}.” The “main proposition of arithmetic” says that “If there is a one to one reciprocal correspondence between {1, 2, ..., m} and {1, 2, ..., p}, then m = p.” Griss then provides seven propositions that hold in this negationless mathematics. Let us go through them.
I a = a
This perhaps follows from the idea that the members of our set are all, by stipulation of their construction, self-same. It seems to be like a reflexive property.
II a = b → b = a
This is a sort of symmetrical property. It would seem to be a contradiction otherwise.
III a = b and b = c → a = c
Here is transitivity, which also makes sense especially in this mathematical context.
IV a ≠ b → b ≠ a
This also would seem to be a contradiction if otherwise.
V a = b and b ≠ c → a ≠ c
This seems to following the notion of equality.
VI a = b or a ≠ b
For this one, Griss writes that it “replaces the negative proposition: Two natural numbers are the same or not, which in non-intuitionistic mathematics holds in virtue of the principium tertii exclusi, but which in intuitionistic mathematics must be proved.” Throughout all this, it was never very clear to me how unequals is not a negative conception. It seems to be because not-equals is defined not as a lack or negation of being equal, but rather as belonging to a different set of numbers (or perhaps, has a distinguishing additional property or sense, like being larger than, as it might be the n′ number or in the set lower than it). In the wording for the negative proposition, two numbers are either the same or they are not the same. This is somehow different from them being either equal or unequal. It is still hard to articulate the distinction, but being unequal is somehow a positive property. Perhaps it can be understood as each unequal number having some additional distinguishing trait. At any rate, the second point is that the negational formulation holds in a classical setting on account of the principle of excluded third, but in intuitionism that does not work. As we know, the law of excluded middle does not hold in intuitionist logic and mathematics (see Priest, Introduction to Non-Classical Logic section 6.2.8, van Stigt’s “Brouwer’s Intuitionist Programme” sections 1.4.1.3 and 1.5.1.2, Mancosu & van Stigt’s “Intuitionistic Logic” sections 4.2.1, and Nolt’s Logics section 16.2, especially 16.2.7 and 16.2.29.) This means, it would seem, that we cannot simply say, on the basis of logic itself, that either it is the case that two numbers are the same or it is not the case that they are the same. Instead we must offer a positive proof that the two numbers are either equal or not-equal, which we did above in section 1.2.1 on the basis of possible set memberships.
VII a ≠ c for each c ≠ b → a = b.
This one we proved above in section 1.2.2. About it Griss says, “VII runs with negation: If it is impossible, that a is not the same as b, then a is the same as b.” I do not know what the “runs with” negation means. As far as I know, it is supposed to be not negational. So I suspect the “runs with” means is not negational but corresponds functionally with the negational formulation, which would say that if it is impossible for two things to not be the same then they must be the same. Proposition VII says something more like if a and b share all the same differences to everything else, then they are the same. Finally Griss next gives four negative propositions regarding sameness and difference that he has replaced in the positive theory. The first is:
different ⇄ not the same.
I am supposing that the ⇄ means biconditional. So perhaps he is saying that formerly, different meant not being the same. Now it means belonging to different parts of a set.
the same ⇄ not different.
Previously perhaps, sameness was understood negationally as not being different. Now it means sharing the same differences.
the same and different exclude one another.
Previously it may have been said that two things are the same, then they cannot be different. Now perhaps, but I am not sure, we are saying that things can be the same or they can be different, with the fact that in all cases they will not be both being a result of their construction and not an assumption about them. I am guessing.
two natural numbers are either the same or different.
This one is very tricky for me, because I do not know how to distinguish it from the proposition from section 1.2.1 above:
Two elements of the set {1, 2, ..., m} are the same or distinguishable.
And also, I do not see the negation in the formulation. Perhaps in that formulation, “different” is understood as “being not the same” but in the non-negational formulation, “distinguishable” is understood as belonging to different parts of a set. I am guessing wildly, so please see the quotation below.)]
We can also formulate these propositions in the following way, though we anticipate the general theory of sets we hope to treat of in a following paragraph.
If we denote the complementary set of the element a of the set {1, 2, ..., m} by A, the complement of A is a and the sum of a and A is {1, 2, ..., m}.
In this connection I mention the so-called main proposition of arithmetic.
If there is a one to one reciprocal correspondence between {1, 2, ..., m} and {1, 2, ..., p}, then m = p.
I do not repeat proofs which have been already given without using the negation.
For the elements of the set {1, 2, ..., m} the following propositions hold now:
I a = a
II a = b → b = a
III a = b and b = c → a = c
IV a ≠ b → b ≠ a
V a = b and b ≠ c → a ≠ c
VI a = b or a ≠ b
VII a ≠ c for each c ≠ b → a = b.
VI replaces the negative proposition: Two natural numbers are the same or not, which in non-intuitionistic mathematics holds in virtue of the principium tertii exclusi, but which in intuitionistic mathematics must be proved.
VII runs with negation: If it is impossible, that a is not the same as b, then a is the same as b.
I briefly enumerate the negative propositions concerning the relations “the same” and “different” which have been replaced in a positive theory.
different ⇄ not the same.
the same ⇄ not different.
the same and different exclude one another.
two natural numbers are either the same or different.
(1132)
Griss, G.F.C. (1946). “Negationless Intuitionistic Mathematics, I,’’ Proceedings of the Koninklijke Nederlandse Akademie van Wetenschappen, 49, 1127–1133.
Journal PDF here:
http://www.dwc.knaw.nl/DL/publications/PU00014659.pdf
Article PDF here:
http://www.dwc.knaw.nl/DL/publications/PU00018278.pdf
Listing of Griss at this journal:
http://www.dwc.knaw.nl/toegangen/digital-library-knaw/?pagetype=publist&search_author=PE00000531
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