For the following questions:
Let \(D\) = a non-empty set of questions, and suppose that each question in \(D\) is either an assignment question or an exam question (but not both).
For all \(x \in D\), let \(E(x) =\) '\(x \textrm{ is an exam question}\)'
For all \(x \in D\), let \(L(x) =\) '\(x \textrm{ is long}\)'
For all \(x,y \in D\), let \(H(x,y) =\) '\(x\) is harder than \(y\)'
INCORRECT! This statement means that there is no assignment question that is short.
CORRECT! For each question, if it is an assignment question, then it is short.
INCORRECT! This statement says that every short question is an assignment question.
INCORRECT! This statement means that there are no short questions and no assignment questions. That is, there are only long exam questions.
INCORRECT!
CORRECT!
INCORRECT!
INCORRECT!
INCORRECT! Check the order of the quantifiers. This answer says that for each assignment question, there is an easier exam question. This does not preclude the possibility that the hardest question is an exam question.
INCORRECT! The main problem with this statement is that for \(\forall y \in D. [(\textrm{ NOT } E(x)) \textrm{ AND } E(y) \textrm{ AND } H(x,y)]\) to be true, it must be the case that all questions y are exam questions (since \(E(y)\) must be true).
INCORRECT! If we pick \(x\) to be an exam question, then the implication \([(\textrm{NOT } E(x)) \textrm{ AND } E(y)] \textrm{ IMPLIES } H(x,y)\) is always true because \((\textrm{NOT } E(x)) \textrm{ AND } E(y)\) is false regardless of what \(y\) is. So, this logic statement is true as long as there is at least one exam question in the domain, even if there is no harder assignment question.
CORRECT! There is a choice for \(x\) such that \(x\) is an assignment question, and, for every choice of \(y\), if \(y\) is an exam question, then \(x\) is harder than \(y\). Unlike (c), this statement ensures that there is at least one assignment question that is harder than every exam question.
EQUIVALENT! Moving the negation rightward, we get \[ \forall x \in D. \textrm{ NOT } [E(x) \textrm{ AND } \forall y \in D. [(\textrm{NOT } E(y)) \textrm{ IMPLIES } H(x,y)]] \] By DeMorgan's Laws, \[ \forall x \in D. [(\textrm{NOT } E(x)) \textrm{ OR } \textrm{ NOT }(\forall y \in D. [(\textrm{NOT } E(y)) \textrm{ IMPLIES } H(x,y)])] \] Namely, for every question \(x\), either \(x\) is not an exam question, or, it is not the case that \(x\) is harder than all assignment questions \(y\).
EQUIVALENT! First, note that answer (c) is equivalent to the given logic statement (click on it to see why). Next, pick the hardest exam question (or pick one if several are equally hard). Call this question \(x\). From answer (c), we know that there is an assignment question (say, \(y\)) that is at least as hard. Therefore, assignment question \(y\) must be at least as hard as every exam question.
EQUIVALENT! Moving the negation rightward, we get \[ \forall x \in D. \textrm{ NOT } [E(x) \textrm{ AND } \forall y \in D. [(\textrm{NOT } E(y)) \textrm{ IMPLIES } H(x,y)]] \] By DeMorgan's Laws, \[ \forall x \in D. [(\textrm{NOT } E(x)) \textrm{ OR } \textrm{ NOT }(\forall y \in D. [(\textrm{NOT } E(y)) \textrm{ IMPLIES } H(x,y)])] \] Moving the negation rightward again, \[ \forall x \in D. [(\textrm{NOT } E(x)) \textrm{ OR } \exists y \in D. \textrm{ NOT }[(\textrm{NOT } E(y)) \textrm{ IMPLIES } H(x,y)]] \] Re-writing the implication using disjunction, \[ \forall x \in D. [(\textrm{NOT } E(x)) \textrm{ OR } \exists y \in D. \textrm{ NOT }[E(y) \textrm{ OR } H(x,y)]] \] By DeMorgan's Laws, \[ \forall x \in D. [(\textrm{NOT } E(x)) \textrm{ OR } \exists y \in D. [ (\textrm{NOT } E(y)) \textrm{ AND } \textrm{ NOT } H(x,y)]] \] Namely, for every question \(x\), either \(x\) is not an exam question, or there exists an assignment question that is at least as hard as \(x\).
LOGICALLY EQUIVALENT! The given logic statement reads: if \(x\) is an exam question and is long, then there exists some question \(y\) that is an assignment question and it is not the case that \(x\) is harder than \(y\).
NOT LOGICALLY EQUIVALENT! The given statement says that, for every long exam question \(x\), there is an assignment question \(y\) such that it is not the case that \(x\) is harder than \(y\). Therefore, it is possible that \(x\) and \(y\) have the same difficulty.
NOT LOGICALLY EQUIVALENT! The given statement only refers to exam questions that are long, so it is possible that a short exam question is the hardest question.
NOT LOGICALLY EQUIVALENT! Assume that there is at least one long exam question \(x\). Then, the given logic statement says that there exists an assignment question \(y\) that is at least as hard as \(x\). Therefore, \(y\) is an assignment question that is not easier than all long exam questions. However, (d) allows the possibility that there exists some long exam question that is harder than all assignment questions. Therefore (d) is not logically equivalent to the given logic statement.
For the following questions:
Let \(A\) be a non-empty set of animals, and suppose that every animal in \(A\) is either a cat or a dog.
For all \(x \in A\), let \(B(x) =\) '\(x \textrm{ is black}\)'
For all \(x \in A\), let \(C(x) =\) '\(x \textrm{ is a cat}\)'
For all \(x \in A\), let \(D(x) =\) '\(x \textrm{ is a dog}\)'
For all \(x,y \in A\), let \(T(x,y) =\) '\(x \textrm{ is taller than } y\)'
Also, suppose that we are given the predicate EQUALS(x,y) which is true when x and y are the same animal, and false otherwise.
ALMOST! This is only one of the correct answers. It says that there does not exist and animal that is both a cat and not black.
ALMOST! This is only one of the correct answers. It says that, for each animal \(x\), if \(x\) is a cat, then it is black.
ALMOST! This is only one of the correct answers. It says that every animal is either black or a dog. Namely, every animal \(x\) sets (\(B(x)\) OR \(D(x))\) to true except for cats that are not black.
CORRECT! Click each of the answers to see why they are correct.
CORRECT! The given statement says that there exists a dog that is not black.
INCORRECT! This is actually the negation of the given statement.
INCORRECT! The given statement says nothing about black dogs.
INCORRECT! One of the above answers is correct.
INCORRECT! This will always evaluate to false since T(x,y) will evaluate to false when \(y = x\).
ALMOST! It depends on how the English statement is interpreted. In many cases, "THE tallest animal" implies that there is a unique tallest animal.
This answer says that, for each cat in the domain, there is a taller dog. This would certainly imply that the tallest animal is not a cat. However, this does not mean that the tallest animal is a dog, because there may not necessarily be a tallest animal! For example, this answer will evaluate to true if there are two or more dogs of the same height that are taller than all cats.
CORRECT! This says that there exists some animal \(x\) such that \(x\) is a dog, and, if \(x\) is not taller than some animal \(y\), then \(x = y\). Unlike answer (b), this implies that there is one animal that is the tallest.
INCORRECT! Only one of the above is correct.
INCORRECT! This statement only says something about black cats and black dogs. It might be the case that the shortest animal is a non-black cat, or a non-black dog, or even a black dog (if there are no black cats).
INCORRECT! This statement only says something about black cats. It might be the case that the tallest animal is a non-black cat.
CORRECT! The statement says that, for all animals \(x\), if \(x\) is a black cat, then, for all animals \(y\), if \(y\) is a black dog, then \(y\) is taller than \(x\). Namely, each black cat is shorter than all black dogs.
INCORRECT! One of the above is correct.
INCORRECT! This answer says that, if \(x\) and \(y\) are black dogs, then \(x=y\). This means that there is at most one black dog, which includes the possibility of zero black dogs.
CORRECT! This answer says that there exists \(x\) that is a black dog, and, if \(y\) is a black dog, then \(y=x\). Therefore, there is at least one black dog, and less than two black dogs.
INCORRECT! One of the above answers is wrong.
INCORRECT! One of the above is correct.