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Logic Axioms | Homework Help Website

Homework 5: Due Thursday October 18 Consider the following Axiom System

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Axiom System IV

The Primitive Terms: player, team, and recruit

Axiom 1: There exists at least one team.

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Axiom 2: Each team recruits exactly two players.

Axiom 3: For each team, there is exactly one other team such that no player is recruited by both.

Axiom 4: Each player is recruited by exactly two teams.

Prove the following theorems.

Theorem 1 There exist exactly four players.

Theorem 2 For each player, there exists at least one other player such that no team recruits both.

Theorem 3 For each pair of players, there exists at most one team recruiting both.

Theorem 4 For each player, there is exactly one other player such that no team recruits both.

 

Learning Activity: Axiom System II

Name:

Primitive Terms: elements, set, contains

Axiom 1: There exists at least one set.

Axiom 2: For each element, there exists exactly one other element such that no set contains both.

Axiom 3: Each set contains exactly two elements.

Axiom 4: Each element is contained in exactly four sets.

Prove Theorem 1: There exist at least eight sets.

Solution:

Proof. By Axiom 1, there exists a set, we can call it s1. By Axiom 3, this set must contain exactly two elements. So it contains some element e1. Now Axiom 4 tells us each element is contained in exactly four set. Thus e1 is contained in three other sets, say s2, s3, and s4. By Axiom 2, there must be some other element e2 such that e2 is not in sets s1, s2, s3 or s4. But by Axiom 4, this element must be in sets s5, s6, s7, s8. So we have shown there are at least eight sets, s1 . . . s8.

Primitive Terms: elements, set, contains

Axiom 1: There exists at least one set.

Axiom 2: For each element, there exists exactly one other element such that no set contains both.

Axiom 3: Each set contains exactly two elements.

Axiom 4: Each element is contained in exactly four sets.

Prove Theorem 2: There exist at least four elements.

Solution:

Proof. By Axiom 1 there exists at least one set, we can call it s1. By Axiom 3, this set must contain exactly two elements, say e1 and e2. Now by Axiom 2, there must be some element e3, such that no set contains e1 and e3. Similarly by Axiom 2, there must be some element e4, such that no set contains e2 and e4. Now e3 and e4 must be different otherwise e4 would have two elements, e1 and e2 that have no sets in common, violating Axiom 2. Thus we have found at least four elements, e1, . . . e4.

Primitive Terms: elements, set, contains

Consider the Axioms:

Axiom 1: There exists at least one set.

Axiom 2: For each element, there exists exactly one other element such that no set contains both.

Axiom 3: Each set contains exactly two elements.

Axiom 4: Each element is contained in exactly four sets.

Prove Theorem 3: For each set, there exist at least two elements not contained in it.

 

Solution:

Proof. Let s1 be an arbitrary set. By Axiom 3, s1 contains two elements, say e1 and e2. Now by Theorem 2, there exists at least two more elements, e3 and e4. Now e3 and e4 cannot be contained in s1, else Axiom 3 would be violated. Thus there are at least two elements not contained in s1. Since s1 was arbitrary, the theorem holds for all sets.

OR

Proof. Let s1 be a set. By Axiom 3, this set contains two elements, say e1 and e2. By Axiom 2, there is exactly one other element, such that there is no set containing it and e1. Since e1 and e2 are both in set s1, this element cannot be e2, we can call it e3. Likewise by Axiom 2, there is exactly one one other element such that there is no set containing it and e2. Since e1 and e2 are contained in s1, that element cannot be e1. It cannot be e3, since e1 and e3 do not share a set. So if e2 and e3 did not share a set, it would violate Axiom 2. Thus this element is some other element, say e4.

We have now found two elements, e3 and e4 that are not contained in s1. Since s1 was an arbitrary set, the theorem holds for all sets.

Quiz 5: Axiom Proofs

Name:

“The way I see it every life is a pile of good things and bad things. The good things don’t always soften the bad things, but vice-versa, the bad things don’t necessarily spoil the good things or make

them unimportant.” Dr. Who

Axiom 1: There exists at least one doctor

Axiom 2: Each doctor travels with exactly two companions

Axiom 3: Every pair of companions travels with one common doctor

Axiom 4: For each doctor, there are exactly two companions that do not travel with him.

Primitive Terms: Companion, doctor, travels

1. (4 points) For every pair of companions, there is at least one doctor that does not travel with them.

Solution: Let c1 and c2 be an arbitrary pair of companions. By Axiom 3, c1 and c2 travel with a common doctor d1. Now by Axiom 4, there are two companions c3 and c4 that do not travel with d1. These cannot be c1 and c2, since d1 travels with them. By Axiom 3, c3 and c4 must travel with a common doctor d2. So we see for c1 and c2 there is a doctor d2 that does not travel with them.

“You know that in 900 years in time and space I’ve never met anyone who wasn’t important before.” Dr. Who

Axiom 1: There exists at least one doctor

Axiom 2: Each doctor travels with exactly two companions

Axiom 3: Every pair of companions travels with one common doctor

Axiom 4: For each doctor, there are exactly two companions that do not travel with him.

Primitive Terms: Companion, doctor, travels

2. (6 points) There are exactly 6 doctors.

Solution: By Axiom 1, there exists at least one doctor, say d1. By Axiom 2, d1 must travel with two companions, c1 and c2. By Axiom 4, there are exactly two companions that do not travel with d1, say c3 and c4. By Axiom 3, c3 and c4 must travel with one common doctor, say d2. Now by Axiom 3, c1 and c3 must travel with a common doctor. This doctor cannot be d1 or d2 else a doctor would travel with more than two companions, violating Axiom 2. So c1 and c3 travel with d3. Likewise there must be a distinct doctor for each pair of companions. Since there are six pairs of companions, we have d4, d5, and d6. So there are at least six doctors.

Suppose there is a seventh doctor d7. By Axiom 2, d7 must travel with two companions. He cannot travel with two of c1, c2, c3 or c4, else Axiom 3 be violated for some pair of companions.

Case 1: Suppose d7 travels with c5 and one of c1, c2, c3 or c4. Without loss of generality, suppose it is c1. Then d7 does not travel with c2, c3 or c4 violating Axiom 4. So d7 cannot travel with c1 and c5.

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Case 2: Suppose d7 travels with c5 and c6. Then d7 does not travel with c1, c2, c3 or c4 contradicting Axiom 4. So d7 cannot travel with c5 and c6

Since those are all possible cases d7 can travel with, d7 cannot travel with two companions, contradicting Axiom 4. So d7 cannot exists. Thus there are exactly six doctors.

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