Theorems 1–8
Theorem 1 (Landau 1). If A ≠ B, then A’ ≠ B’.
Proof: This is a restatement of Corollary 2 of Axiom 2. If Theorem 1 is false, then it is possible for A ≠ B and A’ = B’; but two numbers cannot have the same successor, since then A’ and B’ would be the same number.
Theorem 2 (Landau 2). A’ ≠ A.
Proof: This is a restatement of Axiom 1, since only one number can be formed by the addition of units. This can be shown in a direct way like this: if 1’ = 1, then 1 would be the successor of a number, i.e. 1 = A’ (by Common Notion 2), but that contradicts Corollary 3 of Axiom 2. Therefore, A’ ≠ A is true of 1.
If A’ = A for any given number A, then A would be equal to its own successor, which violates the definition of a successor (Definition 4, which is Corollary 1 of Axiom 2). Therefore, A’ ≠ A is true of 1 and of any given number and so of every number.
Theorem 3 (Landau 4). To every pair of numbers, A, B, the sum of A and B, written A + B, is such that
a). A + 1 = A’, for every A;
b). A + B’ = (A + B)’, for every A and B.
Proof: a).: A + 1 = A’, for every A, follows from the definition of the successor of a number (Definition 4, Axiom 2, Corollary 1) and Common Notion 2.
b).: to prove that A + B’ = (A + B)’, for every A and B, we begin by proving it true of 1 and then of the successors of any given pair of numbers:
A + (1 + 1) = A + 1 + 1 = (A + 1) + 1 = A’ + 1 = (A’)’;
and, since there is no difference between (A’)’ and A’’, we have
(A’)’ = A’’ = A + 1 + 1 = (A + 1) + 1 = (A + 1)’, which is the desired result; as an aside, this is only when B = 1; but 1 could equally well be substituted by A; nevertheless, we also show it here as follows
1 + B’ = 1 + (B + 1) = 1 + B + 1 = (1 + B)’.
A’ + B’’ = (A + 1) + ((B + 1) + 1) = (A + 1) + (B + 1) + 1 = ((A + 1) + (B + 1)) + 1 = (A’ + B’)’.
Therefore, since A + B’ = (A + B)’ is true of 1 and of the successors of any pair of numbers A and B, it is true of all numbers.
Theorem 4 (Landau 5). A + (B + C) = (A + B) + C (Associative Law of Addition)
Proof: Choose any A and B and we shall prove that it holds for all numbers, C. First, we prove it for 1:
A + (B + 1) = (A + B) + 1 is true as proven in Theorem 3. Second, we prove it for the successor of C:
assume that A + (B + C) = (A + B) + C for some C, which C is either 1 or a number; it follows that
A + (B + C’) = A + (B + (C + 1)) = A + B + (C + 1) = A + B + C’ = (A + B) + C’.
Theorem 5 (Landau 6). A + B = B + A (Commutative Law of Addition)
This follows from Common Notion 2 straightforwardly, but to prove it by direct reasoning—since the sum of two numbers can be expressed as additions of as many units as there are in both numbers (and such are only finitely many), for A + B, from the expression as added 1s, we can consider the first B additions of 1 to be B and the remaining A additions of 1 to be A, by which we obtain: A + B = B + A.
Theorem 6 (Landau 7). B ≠ A + B.
This follows from the definition of a number and from Axiom 2, Corollary 2. It is also easy to see by direct reasoning through known properties and theorems:
For A = 1, B ≠ A + B is true simply because A + B is the successor of B and thus not B.
If B ≠ A + B is true of A, then B ≠ A’+ B, since A’ + B = (A + B)’ and the only way for B to be such that B = A’ + B would be for B = (A + B)’ = A + B’, which is impossible, since B cannot be equal to itself, much less equal to itself plus A.
Theorem 7 (Landau 8). If B ≠ C, then A + B ≠ A + C.
This follows from Common Notion 2. But it also can be seen through by direct reasoning through established properties and theorems:
Consider a fixed B and C such that B ≠ C and assume that A + B ≠ A + C holds for A. We see that B’ ≠ C’ and so 1 + B ≠ 1 + C; therefore this is true of A = 1. For the successor of any A for which it holds that A + B ≠ A + C, then
(A + B)’ ≠ (A + C)’, from Theorem 1,
A’ + B ≠ A’ + C, from Theorem 3.b.
Theorem 8 (Landau 9). For any given A and B, one of the following must be the case: 1). A = B; 2). A = B + D, for exactly one D; 3). or B = A + E, for exactly one E.
This follows from the definition of number and from Axiom 1. But it can also be seen as follows:
Cases 1 and 2 and 1 and 3 are not compatible. Cases 2 and 3 are incompatible as well, since otherwise, if they were compatible we would have
A = B + D = (A + E) + D = A + (D + E) = (D + E) + A, which contradicts Theorem 6. This shows that at most one case can be true for all numbers.