quiz3

2.10: A Cycle in a Partial Order? It is possible to have a set A, a partial order R on A, and three distinct elements x, y, and z of A such that the pairs (x, y), (y, z), and (z, x) are all in R.

FALSE (correct) Rationale: If (x, y) and (y, z) are in R, by transitivity (x, z) must be in R as well. But by antisymmetry, (x, z) and (z, x) cannot both be in R, since x ≠ z.

2.9: Composing Functions Let N be the set of naturals. Let d be the function from N to N defined so that d(n) = 2n. Let e be the function from N to N defined so that e(n) = 2n+1. Let m be any positive natural such that m > 1. Then there is exactly one function f from N to N that has f(1) = m and can be formed by composing copies of d and e with one another.

FALSE (correct) Rationale: The first function applied (f) must be one-to-one, since otherwise, two distinct elements from A will be mapped onto one from B, which g will further map onto one element in C, and then g ° f would not be one-to-one. But g need not be one-to-one: consider A = {a, b}, B = {1, 2, 3} and C = {x, y} with f(a) = 1, f(b) = 2, g(1) = x, g(2) = g(3) = y. Since element 3 is not a value of f, g(3) has no bearing on the properties of g ° f.

2.6: Existence and Specification Assuming that Duncan is a terrier, and that all dogs are good, the following proof is invalid. "Since all dogs are good, by Specification, Duncan is good. Since Duncan is a terrier by assumption, he is both good and a terrier. Therefore, by Existence, there exists a dog who is both good and a terrier.

FALSE (correct) Rationale: These are perfectly good usages of Specification and Existence

2.6 Instantiation Let X be a set of dogs including Cardie and Duncan, and F(x) be a predicate meaning "x is fast". If we are given the predicate ∃ x: F(x), where x is of type X, it would be correct to conclude that Cardie or Duncan are fast, but not that both of them are.

FALSE (correct) Rationale: We can conclude that there is at least one fast dog in X, but not anything else about it -- not that it would be Cardie or Duncan. It is also possible that Cardie and Duncan are both fast, but we can't either prove or disprove this from the given statement.

2.3: Consistent Statements? Let A be a set of animals and define two unary predicates over A so that H(x) means "x has hooves" and U(x) means "x is a unicorn". Then it is possible that all three of the following statements are true: (1) "for all x: U(x) → H(x)" (2) "not (for all x: H(x) → not U(x))" (3) "not (exists x: U(x))"

False Rationale: (1) says that all unicorns have hooves, (2) says that it is not the case that all hooved animals are not unicorns, or equivalently that some hooved animal is a unicorn, and (3) says that there are no unicorns. Clearly (2) and (3) contradict one another.

2.10: Another Partial Order Let R be the relation on {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} such that (x, y) is in R if and only if x = y or x > y + 2. Then the Hasse Diagram of R has exactly 18 edges.

False Rationale: 10 has no edge to 9 or 8 because (10, 9) and (10, 8) are not in the relation. It has edges to 7, 6, and 5 but not to 4 or anything smaller because those elements also have edges to them from 4, 5, and 6, but not to 7 or anything larger because these elements also have edges from 6 and so have paths from 10. Thus 10 has three edges going up from it, as do 9, 8, 7, and 6. 5 has edges up to 1 and 2, and 4 has an edge up only to 1, for 18 total edges.

2.10: Hasse Diagram level difference If element a is "path-below" element b in a Hasse diagram, then any upward path from a to b in that diagram must have the same number of lines.

False Rationale: A counterexample has lines for R(a, d), R(d, e), R(e, b), R(a, f), and R(f, b). There are two paths from a to b, one with three lines (a to d to e to b) and one with two (a to f to b).

2.9 One-to-One functions Let f be a function from A to B. If for every element a of A, there is exactly one element b of B such that f(a) = b, then f is a one-to-one function.

False Rationale: If f(a) = f(a'), then f is not one-to-one. Every function has the property that every input has exactly one output, but only some functions are one-to-one.

2.9: Inverse of a Java function Let J be the set of possible Java int values and let f be the function from J to J computed (in real Java) by the method int f (int x) {return x+1;} (If you are not sure how Java handles all possible inputs to this function, look it up.) Then this function does not have an inverse.

False Rationale: It does have an inverse, given by int g (int x) {return x - 1;}. Java does not throw an exception for arithmetic overflow, but just computes, for example, f(MAXINT) as MININT and g(MININT) as MAXINT. So even in the cases where overflow occurs, f undoes the effect of g and vice versa.

2.9 Functions on Strings Let Σ = {a, b} and consider the following functions from Σ* to itself. If w = λ or w ends in a, let f(w) = wb. If w ends in b, let f(w) = wa. Let g(λ) = λ and if w ≠ λ, let g(w) be the string obtained by deleting the last letter of w. Then f and g are inverses of one another.

False Rationale: It is always true that g(f(w)) = w, since f appends a letter and g removes it again. But f(g(w)) need not be w. For example, if w = a, g(w) = λ and thus f(g(w)) = b.

2.10: Minimal Elements A minimal element for a partial order R on a set X is an element a such that for any element b in X, R(b, a) is true if and only if b = a. Then if X is any nonempty finite set, there is exactly one minimal element for R.

False Rationale: It is proved in the text that there must be at least one minimal element. But there can be more than one -- for example, if R is the equality relation every element of X is minimal.

2.8: Is the Relation a Function? Let A be the binary relation on the real numbers defined so that (x, y) is in A if and only if y2 = x. Then A is not a function from the reals to the reals, but A ∩ {(x, y): y ≥ 0} is such a function.

False Rationale: It is true that A is not a function, since it has multiple y values for the same x value. But there is another reason both A and the intersection are not functions -- neither has any y values for negative x values. A function must have exactly one y value for each possible x value.

2.9: Injections and Surjections Let A and B be any two nonempty finite sets with |A| > |B|. Then every function from A to B is a surjection, but none is an injection.

False Rationale: It is true that we cannot have an injection from a larger set to a smaller one. But such a function need not be a surjection, if B has more than one element, because it is possible for all elements of A to be mapped to the same element of B.

2.9: Composition of Functions Let A, B, and C be any three sets. Let f be a function from A to B, g be a function from B to C, and h be the composition g º f, so that h is the function from A to C defined by h(x) = g(f(x)). Then if h is a bijection, both f and g must also be bijections.

False Rationale: Let A = {1}, B = {2, 3}, C = {4}, f(1) = 2, and g(2) = g(3) = 4. Then h(1) = 4, and h is bijection from A to C. But f is not onto, and g is not one-to-one.

2.10 Complements and Partial Orders Let R be a relation on a nonempty set that is a partial order. Define S to be complement of R unioned with the identity relation. That is, (x, y) is in S if and only if either x = y or (x, y) is not in R. Then it is impossible for S to be a partial order.

False Rationale: Let R be a set of numbers with (x, y) in R if and only if x ≤ y. This is a partial order. Then (x, y) is in S if and only if x ≥ y, and this is also a partial order.

2.10: Minimum Elements A minimum element for a partial order R on a set X is an element a such that for every element b, R(a, b) is true. If X is a nonempty finite set, then there must be exactly one minimum element for R.

False Rationale: Let X = {a, b} and let R be the equality relation on X, which we know to be a partial order. Neither a nor b is a minimum element.

2.9 Functions on Infinite Sets Let X and Y be infinite sets and suppose that there exists a function f from X to Y that is both one-to-one and onto. Let g be any function from X to Y. Then g is one-to-one if and only if it is onto.

False Rationale: Let X and Y each be the naturals, let f(x) = x, and let g(x) = x+1. Then g is one-to-one but not onto, while f is both. If X and Y were finite sets, the statement would be true.

2.9 Composition is not onto? Let A, B, and C be three nonempty finite sets, let f be a function from A to B, g be a function from B to C, and h be the composition g º f, so that h is the function from A to C given by h(x) = g(f(x)). Then if |A| < | B|, h cannot be an onto function.

False Rationale: Not necessarily: consider A = {a}, B = {1, 2} and C = {x} with f(a) = 1, g(1) = g(2) = x. g º f is still onto.

2.6 Generalization Let the type of x be "animal", the predicate H(x) mean "animal x has hooves", and the predicate U(x) mean "animal x is a unicorn". The following is a valid proof of the conclusion "All unicorns have hooves" from the premise "it is not the case that there exists an animal that is a unicorn". 1) Let x be an arbitrary animal. 2) By the quantifier DeMorgan rule, the premise is equivalent to "for all y: not U(y)". 3) By Specification, "not U(x)" is true. 4) By the rule of Vacuous Proof, "U(x) → H(x)" is true. 5) Since x was arbitrary, by Generalization we may conclude "for all x: U(x) → H(x)".

False Rationale: Quantifiers can only be introduced at the start of the formula, so the last generalization step is invalid. The statement itself is valid: We note that S(av) for arbitrary v, generalize to ∀ v: SA(av) and then by Existence we conclude ∃ u: ∀ v: SA(uv)

2.10 Complements and Partial Orders Let R be a relation that is a partial order on a nonempty set. Let S be the complement of R, so that (x, y) is in S if and only if (x, y) is not in R. Then it is possible that S is also a partial order.

False Rationale: Since R is reflexive, S must be anti-reflexive, and no relation on a nonempty set can be both reflexive and antireflexive.

2.9: Functions on Java ints Let J be the set of possible Java int values. Then it is possible to define a function f from J to J such that f is onto and f is not one-to-one.

False Rationale: Since the domain and range of this function are the same set, they have the same size. When the domain and range are finite sets of the same size, any function is onto if and only if it is one-to-one.

2.6 Rule of Specification Let T(x) and F(x) denote "dog x is a terrier" and "dog x is fierce" respectively and let d be the constant dog Duncan. From the premises "for all x: T(x) → F(x)" and "T(d)", it would be incorrect to conclude F(d).

False Rationale: Specification gives us T(d) → F(d), and Modus Ponens gives us F(d).

2.10: Complements and Partial Orders Let R be a relation that is antisymmetric. Let S be the complement of R, so that (x, y) is in S if and only if (x, y) is not in R. Then S must also be antisymmetric.

False Rationale: Suppose R is the empty relation on {1, 2}. Then S would contain both (1, 2) and (2, 1), contradicting the definition of antisymmetry.

2.6: Specification Given the premises "All men are mortal" and "Socrates is a man", it is valid to conclude that "All men are Socrates".

False Rationale: The correct conclusion by Specification would be "Socrates is mortal". Woody Allen's character in the film Love and Death uses this argument, when he is attempting to console himself with philosophy while imprisoned.

2.8: Symmetric and Transitive Relations Let A be any set with two or more elements. Then any binary relation on A that is both symmetric and transitive must also be reflexive.

False Rationale: The empty relation is symmetric and transitive but not reflexive. In general, if such a relation contains (x, y) it must also contain (x, x) and (y, y). So it is reflexive if is also satisfies the property that for every x, R(x, y) is true for at least one y. But this property is not guaranteed by the conditions.

2.6: An Existence proof Assume that Duncan is a terrier and that Cardie is a retriever. Then the following proof is valid: "Since Duncan is a terrier, by Existence there exists some dog x who is a terrier. Similarly, by Existence, there exists a dog x who is a retriever. Since x is both a terrier and a retriever, by Instantiation, there exists a dog who is both a terrier and a retriever.

False Rationale: The error is to assign the same variable x to both the terrier and retriever, thereby implicitly assuming that they are the same dog.

2.9 Composition of Onto Functions Let f be a function from A to B and let g be a function from B to C. If the composition g ° f (defined by (g ° f) (a) = g(f(a))) is onto, then both f and g must be onto.

False Rationale: The function g must be onto, since every element of C must be the image under g of some element of B. But f could fail to be onto, as long as every element of C is hit by some element that is hit by f. For example, let A = {a}, B = {b, b'}, and C = {c}, and let f(a) = b), g(b) = c, and g(b') = c. Then g ° f is the function taking a to c, which is both one-to-one and onto, but f itself it not onto.

2.10: Counting partial orders Let X be the set {a, b, c, d}. Consider the set of partial orders P of X in which a is a maximum element (so that P(z, a) is true for any z in X) and d is a minimum element (so that P(d, z) is true for any z). There are exactly two such partial orders.

False Rationale: There are three: P(b, c) might be true and P(c, b) false, P(b, c) false and P(c, b) true, or neither might be true. (They can't both be true because of antisymmetry.) All other values of P are given by the conditions on a and d, aterm-37nd by reflexivity.

2.10: A Minimal Element? Let A be any nonempty set and let R be any partial order on A. Then there must exist a minimal element for R, meaning an element x of A such that for any element y, (y, x) can only be in R if x = y.

False Rationale: This is proved for finite sets in the text. But if A is infinite, it may fail to have a minimal element. For example, if A is the set of all integers under ≤, every element has another element smaller than it.

2.10: Lines in a Hasse Diagram Let n be any natural with n ≤ 4. Let A be a set of n elements and let R be any partial order on A. Then the Hasse diagram of R must have fewer than n lines.

False Rationale: This is true for n = 1 (no lines), n = 2 (no lines or one line), and n = 3 (none, one, or two). But there is a partial order on {a, b, c, d} whose diagram has the four lines (a, c), (a, d), (b, c), and (b, d).

2.6: A Specification proof Assume that there are no unicorns. Then the following proof is invalid: "Assume that it is not the case that all unicorns are pink. Then by the negation-of-quantifier rule, there exists a unicorn that is not pink. By Instantiation, let this unicorn be named Ursula. Since Ursula is a non-pink unicorn, by Separation she is a unicorn, and this contradicts the assumption that no unicorns exist. We reached a contradiction by assuming that it is not the case that all unicorns are pink. Therefore all unicorns are pink."

False Rationale: This proof is essentially what is given as the solution to Exercise 2.6.5 in the text, and is perfectly valid. An almost identical proof demonstrates that all unicorns are non-pink as well.

2.9 Injective and surjective Let A and B be two finite sets, and the statements: P: "Any injection from A to B is also a surjection. "Q: "Any surjection from A to B is also an injection". Then P and Q must be equivalent.

False Rationale: When |A|>|B|, no injections from A to B exist. Therefore, P is vacuously true but Q is false. When |A|<|B|, no surjections from A to B exist. This makes Q vacuously true, but P false. The statements are only equivalent when |A|=|B|.

2.9: Functions from Naturals to Naturals Let f be a function from the naturals to the naturals. Then f is a surjection if and only if it is an injection.

False Rationale: f(n) = n+1 is an injection but not a surjection. g(n) = 0 for n = 0, g(n) = n-1 for n > 0 is a surjection but not an injection.

2.8 Edge relation of a graph Define the relation E on the set {1,2,3,4,5,6,7} so that E(x, y) means "there is an edge from x to y in this undirected graph". The graph has edges (1, 2), (1, 3), (1, 4), (2, 4), (2, 5), (3, 4), (3, 6), (4, 5), (4, 6), (4, 7), (5, 7), and (6, 7). Since it is an undirected graph, all the edges are two-way. Then E is symmetric but not reflexive, antisymmetric, or transitive.

TRUE (correct) Rationale: There are no edges in the graph from nodes to themselves, so E is not reflexive. Every edge is two-way, so E is symmetric. E has two-way edges from one node to another, so it is not antisymmetric. E contains (1, 2) and (2, 5) but not (1, 5), so it is not transitive.

2.6 Rule of Existence Let T(x) and F(x) be predicates on dogs, meaning "x is a terrier" and "x is fierce" respectively. If I am given only the premises "exists x: T(x) → F(x)" and "T(d)", where d is the constant "Duncan", it would be incorrect to conclude that there exists a fierce dog.

TRUE (correct) Rationale: We know that there is some dog that is either not a terrier or is fierce. But if it is not a terrier, we don't know that any fierce dogs are in the set at all.

2.6: Generalization Let the type of x be "animal", the predicate H(x) mean "animal x has hooves", and the predicate U(x) mean "animal x is a unicorn". The following is a valid proof of the conclusion "All unicorns have hooves" from the premise "it is not the case that there exists an animal that is a unicorn". 1) Let x be an arbitrary animal. 2) By the quantifier DeMorgan rule, the premise is equivalent to "for all y: not U(y)". 3) By Specification, "not U(x)" is true. 4) By the rule of Vacuous Proof, "U(x) → H(x)" is true. 5) Since x was arbitrary, by Generalization we may conclude "for all x: U(x) → H(x)".

True

2.8 Inverting a well-defined relation Let X and Y be any two non-empty sets. Let R be any well-defined relation from X to Y. Define a relation S from Y to X by the rule R(x, y) ↔ S(y, x). Then S may or may not be a well-defined relation.

True

2.8 Well-Defined and Total Let X be the set {0, 1, 2}. Then there exist four binary relations on X such that A is both total and well-defined, B is total but not well-defined, C is well-defined but not total, and D is neither total nor well-defined.

True Rationale: A = {(0, 0), (1, 1), (2, 2)}, B = {(0, 0), (0, 1), (1, 1), (2, 2)}, C = {(0, 0)}, and D = {(0, 0), (0, 1)}.

2.10: Total Orders From Partial Orders? Let R be any partial order. Then there exists a total order S such that R is a subset of S.

True Rationale: All we need to do is to add additional pairs to R to resolve the "ties" between any pairs of incomparable elements. The proof of the Hasse Diagram Theorem in the text also shows that this must be the case. You can recursively find this total order by removing a minimal element x, finding a total order that includes all the pairs of R on the remaining elements, then add x back in together with the pairs (x, y) for every y in the set.

1.8: Proof Techniques If I show "(P and (Q → R)) → S" and "(P and (R → Q)) → S", I may conclude "P → S".

True Rationale: Because "not (R → Q)" implies "Q → R", we use the second statement and Hypothetical Syllogism to prove "(P and not (R → Q)) → S". This, with the first statement, gives a Proof By Cases of "P → S".

2.9: Composition of Functions Let A, B, and C be three nonempty finite sets, with |B| < |A|. Let f be a function from A to B, g be a function from B to C, and h be the composition g º f, so that h is the function from A to C given by h(x) = g(f(x)). Then h cannot be a one-to-one function.

True Rationale: Because B is smaller than A, f cannot be an injection, so there must be two elements x and y such that f(x) = f(y). But then h(x) = h(y) as well and h is not an injection.

2.6: Existence Given the premises "Duncan is a terrier" and "Duncan is a dog", it is valid to conclude that "There exists a dog who is a terrier".

True Rationale: Because Duncan is of the correct type by the second premise, we can use the Rule of Existence on the first premise to get the conclusion.

1.7 An equivalence The statements "not (p → (q → r))" and "p and q and not r" are equivalent.

True Rationale: By Definition of Implication and DeMorgan, "not (x → y)" is equivalent to "x and not y". We apply this new rule twice to get the result.

2.8: Inverting a well-defined relation Let X and Y be any two non-empty sets. Let R be any well-defined relation from X to Y. Define a relation S from Y to X by the rule R(x, y) ↔ S(y, x). Then S may or may not be a well-defined relation.

True Rationale: If R is empty, both R and S are well-defined. If R is {(x, y), (x', y)}, it is well-defined but S = {(y, x), (y, x')} is not.

2.10: Hasse Diagram for total order If the Hasse diagram for relation R has two different lines going upward from some element a, then R is not a total order.

True Rationale: If R(a, b) and R(a, c) are both true with b ≠ c, then if R is a total order than either R(b, c) or R(c, b) must be true. But if R(b, c) is true the diagram would have no line from a to c, and if R(c, b) were true it would have no line from a to b. So there can't be two different lines starting at a.

2.10: Hasse Diagrams Let R be a partial order on some set X. Then R is not the equality relation on X if and only if there are one or more edges in the Hasse diagram for R.

True Rationale: If there are no edges, then no two unequal elements are related by R. But each element is related to itself because R is reflexive. So R is the equality relation. If R is not the equality relation, there must be two unequal elements x and y such that R(x, y) is true. But then there must be an upward path in the Hasse diagram from x to y, and this means that the diagram must have one or more edges to form this path.

2.8 Definition of a Function Let I be the set of real numbers {x: 0 ≤ x ≤ 1}. Let R be the binary relation on I defined as {(x, y): x2+ y2 = 1}. Then R defines a function from I to I.

True Rationale: If x is any element of I, there is exactly one value of y, given by y = sqrt(1 - x2), that puts (x, y) in R.

2.9: Two binary relations Let X and Y be any sets, and let R be any binary relation from X to Y. Define S to be a binary relation from Y to X, so that for any x in X and any y in Y, S(y, x) is true if and only if R(x, y) is true. Then S is one-to-one if and only if R is well-defined, and S is onto if and only if R is total.

True Rationale: In each case, switching the roles of x and y in one definition gives exactly the other definition.

2.8: A Relation With Five Properties? Let A be any set with more than one element. Then there exists a binary relation R on A such that all of the following are true: 1) R is symmetric, 2) R is antisymmetric, 3) R is transitive, 4) R is not reflexive, and 5) R's complement is not reflexive (that is, R is not antireflexive)

True Rationale: Let x be any element of A. The relation {(x, x)} has all these properties -- it is not reflexive because A is given to have another element y, and (y, y) is not in the relation.

2.10: Reversal of a partial order Let P be a partial order on some set X. Let Q, the reversal of P, be the relation on X defined so that for any elements a and b of X, Q(a, b) is true if and only if P(b, a) is true. Then Q is also a partial order.

True Rationale: Q(a, a) is always true because P(a, a) is true. If Q(a, b) is true, this implies P(b, a), P(a, b), and then Q(b, a). If Q(a, b) and Q(b, c) are both true, then so are P(b, a) and P(c, b). These latter two facts imply P(c, a) by the transitivity of P, so Q(a, c) is true.

2.6: Instantiation Given the premises "There exists a dog who is a terrier" and "Duncan is a dog", it would be invalid to conclude that "Duncan is a terrier".

True Rationale: Some dog is a terrier, but there is no reason to conclude that this dog is Duncan.

2.8: A relation on strings and letters Let Σ = {a, b} and let R be the relation from Σ* to Σ defined so that R(w, x) means "x is the first character of the string w". Then R is not total, but is well-defined.

True Rationale: The empty string does not have a first letter, so it is not total. But no string has more than one first letter, so it is well-defined.

2.8: Size of a Relation on a Set Let A = {1, 2, 3} and let R be a binary relation on A (a subset of A × A). Then if R is both symmetric and reflexive, its size must be an odd number.

True Rationale: The relation contains the three pairs (1, 1), (2, 2), and (3, 3) because it is reflexive. If it contains (1, 2), it contains (2, 1), and vice versa. Similarly, it contains either both (1, 3) and (3, 1), or neither, and both (2, 3) and (3, 2), or neither. So the number of pairs in the relation could be 3, 5, 7, or 9, all odd numbers.

2.8: Transitive and Reflexive relations Let A be any set with two or more elements, and let R be a binary relation on A that is transitive and satisfies the property "for all x: exists y: R(x, y)". Then R is not necessarily reflexive.

True Rationale: The relation {(a, b), (b, b)} is transitive and satisfies the given property, but is not reflexive as it lacks (a, a).

2.8 Edge relation of a graph Define the relation E on the set {1,2,3,4,5,6,7} so that E(x, y) means "there is an edge from x to y in this undirected graph". The graph has edges (1, 2), (1, 3), (1, 4), (2, 4), (2, 5), (3, 4), (3, 6), (4, 5), (4, 6), (4, 7), (5, 7), and (6, 7). Since it is an undirected graph, all the edges are two-way. Then E is symmetric but not reflexive, antisymmetric, or transitive.

True Rationale: There are no edges in the graph from nodes to themselves, so E is not reflexive. Every edge is two-way, so E is symmetric. E has two-way edges from one node to another, so it is not antisymmetric. E contains (1, 2) and (2, 5) but not (1, 5), so it is not transitive.

2.10 Hasse Diagrams Let D be the division relation (as defined in the text) on the numbers {1, 3, 5, 7, 9, 11, 13, 15}. Then the Hasse diagram for D has eight nodes and eight edges.

True Rationale: There is a node for each of the eight numbers, and lines from 1 to 3, 1 to 5, 1 to 7, 1 to 11, 1 to 13, 3 to 9, 3 to 15, and 5 to 15.

2.10: A Partial Order? Let X be the set {a, b, c, d}. Then the following relation R is a partial order on X: R = {(a, a), (a, c), (a, d), (b, b), (b, c), (c, c), (d, d)}

True Rationale: This is clearly reflexive. It is antisymmetric as none of the three pairs with two different elements have the opposite pair present. And it is transitive as the only two-step paths involve one of the (x, x) pairs.

2.6 Rule of Generalization Let Σ = {a, b} and let all variables be of the type "string over Σ". Let EA(w) mean "string w ends in a". Then the following is a valid proof of the statement "for all u: exists v: EA(uv)": "Let u be an arbitrary string. Let v = 'a'. Then uv ends in a. By Existence, we have proved "exists v: EA(uv)". Since u was arbitrary, we have proved "for all u: exists v: EA(uv)"."

True Rationale: This is totally valid.

2.8: Size of a Relation Let A and B be any two nonempty finite sets. Let R be a relation from A to B. Then if R is a function from A to B, its size is the same as the size of A.

True Rationale: This is true because there is exactly one pair in the relation for each element a of A: the pair (a, f(z)).

2.10 Proof of the Hasse Diagram Theorem Let R be a partial order on a nonempty set. Suppose that in constructing the Hasse diagram of R according to the proof in the text, at some point I have a choice of more than one minimal element. Then R is not a total order.

True Rationale: Total orders have only one minimal element, since if x and y are any two distinct elements then either (x, y) or (y, x) is in the relation, so that x and y cannot be minimal. If we started with a total order, it would remain a total order as we keep removing and drawing minimal elements.

2.10: The Division Relation Let D be the division relation on naturals, so that (x, y) is in D if and only if there exists a natural z such that xz = y. Then if u and v are any two naturals, there exists a natural w such that (u, w) and (v, w) are both in D.

True Rationale: We can take w to be uv, and then (u, uv) and (v, uv) are both clearly in D. It doesn't matter if either u or v are 0.

1.4: Completeness of operators It is not possible to write all possible compound boolean propositions using only the → operator and atomic variables. (Thus the set {→} is not complete in the sense of Problem 1.4.8.)

True Rationale: We cannot write the constant false! If all the atomic variables are true, then any expression combining atomic variables with → must be true, since 1 → 1 is equivalent to 1. But if the expression were equivalent to 0, it would be false for any assignment of the atomic variables.

2.6 Rule of Instantiation Let X be a set of dogs including Duncan, and let T(x) mean "x is a terrier". If we are given only the premise "exists y: T(y)", where the type of the variable y is X, it would be incorrect to conclude that Duncan is a terrier.

True Rationale: We know that some terrier exists, but not that it is Duncan.

2.6 Rule of Existence Let T(x) and F(x) be predicates on dogs, meaning "x is a terrier" and "x is fierce" respectively. If I am given only the premises "exists x: T(x) → F(x)" and "T(d)", where d is the constant "Duncan", it would be incorrect to conclude that there exists a fierce dog.

True Rationale: We know that there is some dog that is either not a terrier or is fierce. But if it is not a terrier, we don't know that any fierce dogs are in the set at all.

2.8 Neither well-defined nor total Let X be the set {0, 1}. Then there exist four binary relations on X such that A is both total and well-defined, B is total but not well-defined, C is well-defined but not total, and D is neither total nor well-defined.

True Rationale: We may let A = {(0, 0), (1, 1)}, B = {(0, 0), (0, 1), (1, 1)}, C = {(0, 0)}, and D = {(0, 0), (0, 1)}.

2.6 Existence Proof Let T(x) denote "x is a terrier", and F(x) denote "x is fast", with all variables and predicates over a set X of dogs that includes a dog d. Then, from premises T(d) and ∀ x: F(x) we can derive ∃ x: F(x) ∧ T(x)

True Rationale: Yes, from ∀ x: F(x) we can derive F(d) by specification, and from F(d) ∧ T(d) we apply the rule of Existence.