Question text is in black, answers in blue.
You have two messy combinations of AND's and OR's that you want to show to be equal. In algebra, one thing you might try with two polynomials is to put them each in a normal form, with similar terms combined and terms in order by degree, because it's easy to tell whether two polynomials are equal if each one is in this form. Here, you might find it useful to put the two compound propositions into conjunctive normal form (an AND of clauses, each of which is an OR), or disjunctive normal form (an OR of clauses, each of which is an AND). With two compound propositions each in the same one of these forms, you can use the commutative and associative rules for AND and OR to turn one of them into the other. This would be an equational sequence proof because you are always turning one compound proposition into another that is equivalent.
Sorry, I should have defined this term in Section 1.1.
Remember that if A ⊆ B, every element of A is an element of B
(∀x: (x ∈ A) → (x ∈ B)) but the sets A and B might or
might not be equal. A is a proper subset of B (sometimes written
"A ⊂ B" instead of "A ⊆ B") if A is a subset of B but the sets are
not equal. For this to happen, there must be an element of A that is
not an element B: ∃x: (x ∈ B) ∧ (x ∉ A).
For each of these for problems, you need to either give a quantifier
proof of the given statement from the premises given (including the two
statements above for "proper subset", or come up with an example of sets
A and B, and predicates P and Q, such that all the premises are true but the
given implication is false.
By convention, a Java variable written in ALL CAPS
is a constant. This method you are writing would live in a class,
where the public static final int variable ALENGTH has been
set to something. If it were 100, for example, this set A would be a
subset
of the set {0,1,2,...,99}.
Yes, as it says on page 1-9 of the text, "" in Java is the same thing
as λ in the notation of this course. So in this Java context,
isInA("") is a boolean that is true if the
empty
string is in A and is false if it is not.
You don't. The compound proposition doesn't change at all. The only thing that changes is that now your truth table has 16 lines instead of 8, because it is a truth table where each line has a setting for four variables instead of three. It's true that the value of s doesn't affect the value of the compound proposition, so that the line for "0 0 0 0" and the line for "0 0 0 1" will be the same (if the variables are in the order "p q r s". From this fact you should be able to answer the question.
Yes, you're right. Please add in the ")" before the AND sign in the fourth line.
Last modified 10 February 2011