=========================================================================== CSC 165H Lecture Summary for Week 5 Winter 2004 =========================================================================== ----------------- Mixed Quantifiers ----------------- In general, \-/ y, -] x, p(x,y) not equivalent to -] x, \-/ y, p(x,y). In particular, (*) -] x, \-/ y, p(x, y) implies (**) \-/ y, -] x, p(x, y) but not the other way around (a single x works for (*) and can be used for each y in (**)). People sometimes emphasize that x in (2) is allowed to depend on y, by subscripting it: \-/ y, -] x_y, p(x_y, y). This suggests that x is "a function of y". In everyday language: For every y there's a _corresponding_ x such that p(x, y). ----------------------- -> in terms of ~ and \/ ----------------------- Consider general diagram for two sentences P and Q, and possible regions for P -> Q. Can we express this as the union of two of the regions for P, Q, ~Q, ~P? [Draw diagrams.] Yes: ~P \/ Q. What about ~ (P -> Q)? Either directly from diagrams, or through sequence of equivalences: ~ (P -> Q) <=> ~ (~P \/ Q) <=> ~~P /\ ~Q <=> P /\ ~Q (using DeMorgan's laws). ------------------------------------ \-/ (P \/ Q) vs (\-/ P) \/ (\-/ Q) ------------------------------------ Consider: (1) Every employee makes more than 30K, or every employee makes less than 50K. To check this, we check the truth of: Every employee makes more than 30K. Every employee makes less than 50K. But is this the same as: (2) Every employee makes more than 30K or less than 50K. (2) is true but (1) is false. (2) gives us choice that can depend on the employee. (1) requires a choice that works for all of them. In general, (\-/ x, p(x)) \/ (\-/ x, q(x)) implies \-/ x, (p(x) \/ q(x)). But the converse doesn't have to be true. ------------------------------------ \-/ (P /\ Q) vs (\-/ P) /\ (\-/ Q) ------------------------------------ Consider now: (3) Every employee makes more than 30K, and every employee makes less than 50K. (4) Every employee makes more than 30K and less than 50K. (3) and (4) are equivalent: unlike \/, there's no choice involved. (\-/ x, p(x)) /\ (\-/ x, q(x)) is equivalent to \-/ x, (p(x) /\ q(x)). --------------------------------- -] (P /\ Q) vs (-] P) /\ (-] Q) --------------------------------- How about with existence? (5) There's an employee who makes more than 25K, and there's an employee who makes less than 10K. (6) There's an employee who makes more than 25K and less than 10K. In (5), there are two existence statements, with each one's "employee" local to it. They don't have to be the same employee. Anya proves the first half of (5), and Dawn proves the second half. So (5) is true. In (6), there is one existence statement. We need to find an employee with both properties. There is none, so (6) is false. In general, -] x, (p(x) /\ q(x)) implies (-] x, p(x)) /\ (-] x, q(x)), because the second statement lets us choose x based on whether we want it to satisfy p or q. But the converse might not be true. --------------------------------- -] (P \/ Q) vs (-] P) \/ (-] Q) --------------------------------- Consider now: (7) There's an employee who makes more than 60K, or there's an employee who makes exactly 7K. (8) There's an employee who makes more than 60K or exactly 7K. These are both false. In general, (-] x, p(x)) \/ (-] x, q(x)) is equivalent to -] x, (p(x) \/ q(x)). -------------------------- Negating mixed quantifiers -------------------------- What does it mean for \-/ x, -] y, p(x, y) to be false, i.e., what is the meaning of: ~ \-/ x, -] y, p(x, y). ...it means there is an x for which there is no corresponding y: -] x, ~ -] y, p(x, y). ...it means there is an x such that all the ys don't work: -] x, \-/ y, ~ p(x, y). As the ~ moves inside, it changes universal to existential and vice versa. What does it mean for -] y, \-/ x, p(x, y) to be false? ~ -] y, \-/ x, p(x, y) \-/ y, ~ \-/ x, p(x, y) \-/ y, -] x, ~ p(x, y) In everyday language: every y fails to work for some x. ----- Proof ----- A proof is a convincing argument that a statement is true. It may build on previously proven statements. To get the process started, proofs of statements about a domain D need a set of basic truths: the statements that in a sense define D, called axioms, that we simply agree are true without proof. For example, if D is the set of real numbers then we have the basic rules about arithmetic and inequalities. --------------------------- Direct proof of implication --------------------------- Consider statements of the form: (1) \-/ x in D, p(x) -> q(x). Many of the statements already known to be true will be in a form similar to (1). Our hope is to find among them: (2.0) \-/ x in D, p(x) -> r1(x). (2.1) \-/ x in D, r1(x) -> r2(x). ... (2.n) \-/ x in D, rn(x) -> q(x). These together prove (1). People like a flexible way to summarize that (2.0)...(2.n) prove (1). They write the proof: Let x in D be such that p(x). Then r1(x) (by 2.0). So r2(x) (by 2.1). ... So q(x) (by 2.n). This form emphasizes what an existing result adds to our understanding. And when it's obvious which result was used, we can easily avoid mentioning it. This form seems to talk about only one x, but by not assuming anything more about about x than x in D and p(x), it applies to all x in D with p(x).