Exercise 1.2. Consider the Following Axiom Set

Exercise 1.2. Consider the following axiom set. A1. Every hive is a collection of bees. A2. Any two distinct hives have one and only one bee in common. A3. Every bee belongs to two and only two hives. A4. There are exactly four hives. T1. There are exactly six bees.

Proof: [note: Axioms reworded: A1. There are things called hives and bees, and bees are things that are in hives. Hives contain only bees. A2. If there are 2 hives, then there is exactly one bee that is in both hives A3. If there is a bee, then there are exactly two hives that it belongs to A4. There exist exactly 4 hives

There are no hypotheses for this theorem beyond the axioms, so we must start with one of the axioms.

A1 doesn’t say much useful, it mostly introduces the undefined terms and excludes other random things besides bees from being in hives

A2 and 3 have if’s, and we don’t know anything yet, so we can’t use those. A4 tells us something without any ifs, so we need to start there.]

(1) By A4, we know there are exactly 4 hives, call them H, I, J, K.

[now we have some hives, so we can use A2]

(2) By A2, there is exactly one bee that is in H and I, call it a.

[now we have a bee, so we can use A3]

(3) By A3, a belongs to only two hives, H, and I, so it can’t belong to J or K

[repeat for other combinations of hives]

(4) By A2, there is exactly one bee that is in H and J. By line (3), that bee cannot be a because a is not in hive J, so this must be a new bee, call it b

(5) By A3 there b belongs to only two hives, so it does not belong to hives I and K (6) By A2, there is exactly one bee that is in H and K. By line (3), that bee cannot be a because a is not in K , and it cannot be b by line (5) because b is not in K, so this must be a new bee, call it c.

(7) By A3, c cannot be in I or J (8) Similarly, there exist bees d, e, f that are distinct from bees a, b, c and from each other, such that  d is in I and J and not H or K  e is in I and K, and not H or J  f is in J and K and not H or I.

[note, this use of similarly is allowed after the pattern for proving each step is well established.]

(9) Therefore, there exist at least 6 bees.

(10) Suppose there exists another bee, call it g.

[this is the beginning of a proof by contradiction]

(11) By A3, g belongs to two hives.

(12) Suppose that g belongs to hives H and I

[beginning of a sub-proof by contradiction]

(13) by A2, there is only one bee in H and I, but we have by lines (2) and the assumption (12) that both a and g are in hives H and I which is a contradiction [contradiction]

(14) Thus, the assumption must be false, and g is not in H and I. [the contradiction negates the most recent assumption, which is line 12]

(15) Similarly, we can show that g does not belong to the pairs H and J, H and K, I and J, I and K or J and K [we can use “similarly” here because the proofs of these statements would be identical to lines 12 and 13 with the obvious substitutions made] (16) There are only 6 pairs of hives, if we have 4 hives to choose from (this can be substantiated by referring to other discrete math results: 4C2=6 or by listing: H can be paired with I, J, or K, all of which are on the list, I can be paired with H, J or K, J can be paired with H, I or K, K can be paired with H, I or J, all of which pairings have been considered), and g cannot be in any of the pairs,

(17) line 16 contradicts A3. [contradiction]

(18) Thus, the seventh bee cannot exist [the contradiction negates the most recent assumption. Assumption 12 was previously negated, so we go back to assumption 10]

(19) Thus there are fewer than 7 bees. (20) by lines (19) and (9) there exist exactly 6 bees. Exercise 1.2. Consider the following axiom set. A1. Every hive is a collection of bees. A2. Any two distinct hives have one and only one bee in common. A3. Every bee belongs to two and only two hives. A4. There are exactly four hives. T2. There are exactly three bees in each hive. [Alternate wordings: Axioms reworded: A1. There are things called hives and bees, and bees are things that are in hives. Hives contain only bees. A2. If there are 2 hives, then there is exactly one bee that is in both hives A3. If there is a bee, then there are exactly two hives that it belongs to A4. There exist exactly 4 hives T2: If there is a hive, X, then there are exactly 3 bees in X] (1) Let X be a hive [the if part of T2]

(2) by A4, there are 3 other hives, call them A, B, C. [I know from the previous proof that bees are associated with pairs of hives, so I’m setting up to get pairs of hives]

(3) by A2, X and A have exactly one bee in common, call it a.

(4) by A3, a does not belong to hives B and C [because a belongs to only two hives: A and X]

(5) by A2, X and B have a bee in common, and that bee cannot be a because a does not belong to hive B, so call the new bee b

(6) by A3, b does not belong to A or C

(7) by A2, X and C have exactly one bee in common, and that bee cannot be a or b, because they don’t belong to C, so call the new bee c.

(8) by (3), (5) and (7), there are at least 3 bees in X. (9) suppose there is another bee d in X [setting up for a contradiction]

(10) by A3, d must belong to another hive besides X:

(11) d cannot belong to hive A, because A and X already have a bee in common, and by A2, they cannot have another bee in common.

(12) Similarly, d cannot belong to hive B or C, which contradicts (10) [here is the contradiction]

(13) Thus, X cannot have another bee. [negates most recent assumption (9)]

(14) So, X has exactly 3 bees (lines 8 and 13) Exercise 1.2. Consider the following axiom set. A1. Every hive is a collection of bees. A2. Any two distinct hives have one and only one bee in common. A3. Every bee belongs to two and only two hives. A4. There are exactly four hives. T3. For each bee there is exactly one other bee not in the same hive with it.

[If-then wording: Axioms reworded: A1. There are things called hives and bees, and bees are things that are in hives. Hives contain only bees. A2. If there are 2 hives, then there is exactly one bee that is in both hives A3. If there is a bee, then there are exactly two hives that it belongs to A4. There exist exactly 4 hives

T3: If there is a bee x, then there is exactly one other bee y such that there is no hive that contains both x and y]

Proof: (1) Let x be a bee.

(2) by A3, x belongs to exactly two hives, call them A and B

(3) by A4, there are exactly two more hives besides A and B, call them C and D

(3a) By A3, since x belongs to A and B, x does not belong to C or D.

(4) by A2, there is exactly one bee in C and D. x is not in C or D, so this bee cannot be x, call this new bee y.

(5) by A3, y cannot belong to A or B because it belongs to C and D.

(6) Of the 4 hives A, B, C, D, A and B contain x but not y and C and D contain y but not x, thus there exists a bee y such that there is no hive that contains both x and y

(7) Suppose there is another bee z such that there is no hive that contains both x and z [setting up for a contradiction] (8) Then z cannot be in A or B, because A and B contain x (by assumption 7)

(9) z must be on two hives, by A3. The only two hives besides A and B are C and D, so z must be in C and D.

(10) (9) and (4) together say that there are two bees in hives C and D, which contradicts A2. [contradiction]

(11) Thus, z cannot exist, and there is no other bee that satisfies the conditions

(12) Therefore, there is exactly one bee y such that there is no hive that contains both x and y (6 and 11)