Mengenlehre I:	Frage 1: On (17), I was wondering if it would also be possible to say that "G is not a superset of A" and "A is not a subset of G"? Antwort: This is correct. The reason why these cases are not mentioned here, perhaps, is that these relations are excluded for rather obvious reasons: if set A has more elements (members) than set G, then we know that A cannot be a subset of G, and that G cannot be a superset of A, even without inspecting the individual elements of A and G. Frage 2: On (27)-(28), I noticed that none of the given answers include a proper superset (such as "C is a proper superset of B"). Is that simply due to space saving concerns here, or is a proper superset symbol not commonly used? Or something else entirely that I've failed to realize? Antwort: There is no deep reason for this. As you say, one could have equally stated that C is a proper superset of B. It's just that once you have stated that B is a proper subset of C, it follows that C is a proper superset of B (and vice versa). Frage 3: On (29), are the following answers also possible: "G is a superset of H", "H is a superset of G", "G is not a proper superset of H", and "H is not a proper superset of G"? ? Antwort: Yes, these statements are equivalent to those given on the slides.
Mengenlehre III:	Frage 1: I didn't really understand the last question of excercise 6 of Excercises 3. We basically have to create the complement of some relation R, and the formula for that would be R' = ((A × B) − R), and we would simply take R−1 instead of R. But what is the base now? What do we use here in place of A × B? Antwort: It was asked to create (R−1)', which is the complement of the inverse of R. R is a relation from A to (A ∪ B) (i.e. R ⊆ (A × (A ∪ B))). The inverse of R (R−1) is also a subset of (A × (A ∪ B)), simply with all the first and second members of each tuple of R flipped. And the complement of this inverse ((R−1)') is then (A × (A ∪ B)) − R−1 (i.e., all the tuples of (A × (A ∪ B)) except those that are members of the inverse of R. Similarly for (R')−1 (the inverse of the complement of R). In this context, note that the solution for (R−1)' that was given on the solution sheet is not correct. Correct is (R−1)' = {〈 b,c 〉, 〈 b,2 〉, 〈 b,3 〉, 〈 c,c 〉, 〈 c,b 〉, 〈 c,2 〉, 〈 c,3 〉} Frage 2: The proofs for ((A − B) + (B − A)) = A + B and for (A + B) ⊆ B iff A ⊆ B went pretty much over my head. I understood them once I saw the solution, but I couldn't have done that on my own - I have no clue which law would be best to use when, it's pretty overwhelming for me. Do you have any tips? Antwort: This is a good question. The answer is that there is, unfortunately, no general recipe as to how to proceed. One just has to do many proofs to develop a feeling as to which law may lead towards the desired simplification. The purpose of the excercise was that you try to solve it, and thereby get some practice. And if you are able to follow the proof once you have been given it, then this means that you are on a good way. And do not worry: you will not be confronted with a proof in the Abschlussklausur that is as complicated as the ones in the exercises.
Relationen und Funktionen:	Frage 1: My question is about the end of the fourth presentation (Relations and functions), to be exact slide 27. I don't understand anti-symmetry. In the example R3, it is said that <b,b> is already its own reverse, so b=b, and that means that it's anti-symmetric. But in R5 <b,b> suddenly isn't anymore the reverse of itself, because it says that R5 is non-symmetric since there are no reverses, but at the same time it is anti-symmetric with the explanation that b=b??? What is happening there? Antwort: Essentially, the answers are correct, but you are right: the justifications given do not make sense in the case of R5. The wording should rather be something like the following: "R5 is non-symmetric (the pair 〈a,c〉 doesn't have a reverse) and also anti-symmetric (the only pair for which there is a reverse is of the reflexive type: 〈b,b〉)" Frage 2: A question on the first excercise of sheet 4: "Let A={a, b, c} and B={1, 2}. How many distinct relations are there from A to B?" The solution given begins with: "Each member of the three members of A can relate to one, two, three or to none of the members of B." I'm confused about the "three", because B only has two members. Antwort: The wording given in the solution was misleading. Meant was: R1 = {〈 a,1 〉, 〈 b,... 〉, ...}, R2 = {〈 a,2 〉, 〈 b,... 〉, ...}, R3 = {〈 a,1 〉, 〈 a,2 〉, 〈 b,... 〉, ...}, R4 = {〈 b,... 〉, ...}, etc. In prose: a can relate only to 1, or only to 2, or to 1 and 2 (that's the third possibility) or to nothing. Frage 3: I've got a question about the last part of Exercise 2. In the solution to R2-1∘R2, why are there those pairs with the 4 in the first position of the tuples? And at the same time: Where are 〈1,1〉, 〈2,2〉 and 〈3,3〉? Antwort: You are completely right. The solution given was not correct. Thank you for pointing this out. Frage 4: In Exercise 4, why is "brother of" non-symmetric? I get that it's not symmetric because not everybody has/is a brother. Antwort: If a relation is not symmetric (i.e. if the definition for symmetry is not fulfilled), then it is called non-symmetric. The relation "brother of" is non-symmetric because if x is male and y female, then x may be the brother of y (i.e. 〈x,y〉 is part of the relation) but not vice versa. Frage 5: Could you give an explanation why "brother of" is not anti-symmetric? Do you have another way of explaining it than in the presentation? And why, in exercise 5, is R1 anti-symmetric? Is it because of the pairs 〈1,1〉, 〈4,4〉 and 〈3,3〉? Shouldn't there be also a pair 〈2,2〉 to call it anti-symmetric? Isn't it necessary for anti-symmetry that every pair 〈x,y〉 must have a pair 〈y,x〉? Antwort: "brother of" is not anti-symmetric because it is not reflexive: x can never be the brother of x. And R1 in exercise 5 is anti-symmetric because all symmetric pairs that are part of the relation are reflexive pairs: 〈1,1〉, 〈3,3〉, 〈4,4〉. Frage 6: What is the exact difference between symmetric and anti-symmetric? Antwort: For symmetry to hold, for every pair in the relation its inverse must also be part of the relation. For anti-symmetry to hold, all the pairs for which the inverse is also part of the relation must be reflexive pairs (i.e., pairs of the type 〈x,x〉).
Aussagenlogik:	Frage 1: In questions 1b. and 2e., we are given the following answer format as a possibility for expressing an 'exclusive or': ((p∨q) ∧ (¬ (p∧q))) I'm hoping you can explain how this works because to me, it feels like the right half would negate the left half; if I have (an apple or an orange) and (not (an apple or an orange)), do I have anything? Do I have both? It seems that I would have nothing, so I think I'm just not understanding the logic behind it (no pun intended). Antwort: First of all, p and q stand for full propositions. Thus, i don't know how far the comparison with apples and oranges will bring us. Second, the right half is not simply the negation of the left one: on the left hand, you have a disjunction, on the right one a (negated) conjunction. The left half is true when a) p is true or b) q is true or c) both are. The right half is only true if not both p and q are both true. In other words, the right half elimnates option c) for truth of the left half. This means that the whole statement (a conjunction) is only true if either p is true or q (but not if both are). This is the semantics of the exclusive "or". Frage 2: In question 1h, (/The stock market advances when public confidence in the economy is rising and only then/), I'm not sure why the answer would be biconditional. The stock market advances when public confidence is rising, but it doesn't seem to be said or implied anywhere that public confidence only rises when the stock market advances. Am I just misunderstanding the biconditional maybe? Antwort: The part up to the "and" means that if public confidence is rising, then the stock market advances. Now, there might, in principle, be other circumstances under which the stock market advances. The "and only then" passage eliminates this option. But this, in turn, means that if the stock market advances, then it must also be the case that the public confidence is rising, which is the reverse of the first implication. Hence, a biconditional.