Solution to Midterm Review Problem - Logical Equivalence | MATH 8, Exams of Mathematics

Download Solution to Midterm Review Problem - Logical Equivalence | MATH 8 and more Exams Mathematics in PDF only on Docsity! Math 8 - Solutions to Midterm Review Problems Winter 2007 1. Prove the logical equivalence: (P∧ ∼ Q) ∧ (R ⇒ Q) ≡∼ [(P ⇒ Q) ∨R]. Solution. Starting with the left-hand side and using the identity A ⇒ B ≡∼ A ∨ B and then the distributive law, we have (P∧ ∼ Q) ∧ (R ⇒ Q) ≡ (P∧ ∼ Q) ∧ (∼ R ∨Q) ≡ (P∧ ∼ Q∧ ∼ R) ∨ (P∧ ∼ Q ∧Q) ≡ (P∧ ∼ Q∧ ∼ R) ∨ F ≡ P∧ ∼ Q∧ ∼ R ≡ ∼ [∼ (P∧ ∼ Q) ∨R] ≡ ∼ [(∼ P ∨Q) ∨R] ≡ ∼ [(P ⇒ Q) ∨R]. Alternatively, we could check that the left and right hand sides have identical truth tables: each side is False only when P is True and Q and R are both False. Another method would be to check that the proposition obtained by changing the “≡” to a “⇔” is a Tautology, by means of a truth table. 2. Simplify the sentential form (P∧ ∼ Q) ⇒ (P ∨Q) as much as possible. Solution. (P∧ ∼ Q) ⇒ (P ∨Q) ≡ ∼ (P∧ ∼ Q) ∨ (P ∨Q) ≡ (∼ P ∨Q) ∨ (P ∨Q) ≡ ∼ P ∨ P ∨Q ∨Q ≡ Q. 3. Write the following propositions symbolically with no words. (You do not have to prove them.) (a) “There does not exist a largest real number.” Solution. ∀x ∈ R ∃y ∈ R (y > x) or a more literal version would be ∼ [∃x ∈ R ∀y ∈ R (y ≤ x)]. (b) “The interval strictly between any two distinct real numbers contains at least one rational number.” Solution. ∀x ∈ R ∀y ∈ R [(x < y) ⇒ ∃z ∈ Q (x < z < y)] 1 (c) “Every nonempty set has at least two distinct subsets.” Solution. (We must assume that some set U of sets is given for the domain of interpretation.) ∀A [(A 6= ∅) ⇒ ∃B ∃C (B 6= C ∧B ⊆ A ∧ C ⊆ A)] 4. Determine whether the following statements are true or false, where the universe of discourse is the set of all real numbers, and give a brief justification. (a) ∀x ∃y [(y > 0) ⇒ (xy > 0)] Solution. True. Notice that the implication is automatically true whenever y ≤ 0. So for any x, one such y that makes the implication true is y = 0. (b) ∀x ∃y ∀z [(x + y)z2 ≤ 0] Solution. True. Since z2 ≥ 0 for any z, the inequality will be satisfied if and only if x + y ≤ 0. So for any x, we can choose y = −x (or any y < −x) to make the inequality true for all z. (c) ∃x ∀y (xy = 1) Solution. False. This says that there is some x that is equal to 1/y for every y 6= 0. Clearly that is impossible. (d) ∀y ∃x (x < y < x + 1) Solution. True. Any y satisfies the inequality y − 1/2 < y < y + 1/2, so we can take x = y − 1/2. 5. Recall that the Sheffer stroke of two propostions P and Q is defined as P ↑ Q ≡∼ (P ∧Q). If A = {x | P (x)} and B = {x | Q(x)}, let S = {x | P (x) ↑ Q(x)}. (Assume everything is contained in a fixed domain of interpretation U .) (a) Describe the set S in terms of A and B, using the standard set operations (eg. union, intersection, set difference, etc.). Solution. S = {x | P (x) ↑ Q(x)} = {x | ∼ (P (x) ∧Q(x))} = {x | P (x) ∧Q(x)}′ = (A ∩B)′. (b) Illustrate S using a Venn Diagram. Solution. Everything should be shaded except for the intersection of A and B. (c) If we also know that A ⊆ B, what else can we say about S? Solution. If A ⊆ B, then A ∩ B = A (think of a Venn diagram, if you are not sure about this). Thus S = (A ∩B)′ = A′ is the complement of A. 2