Due date: Friday, January 27, 2023 10:00 pm (Eastern Standard Time)
Our table of logical equivalences (table 6 in section 1.3) only lists two distributive laws, but the laws $\boxed{(p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r)}\text{ and }\boxed{(p ∧ q) ∨ r ≡ (p ∨ r) ∧ (q ∨ r)}$ are also true, and are also sometimes referred to as distributive laws. Use equivalences that are in the tables to show that these two extra equivalences are true.
$\text{Hint: It might help you to think about the corresponing rules from arithmetic.}\\ \text{ How are the statements }\boxed{a × (b + c) = a × b + a × c }\text{ and }\boxed{(a + b) × c = a × c + b × c}\text{ related?}$
$$ \boxed{(p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r)} $$
<aside> 💡 Proof:
$$ \text{define that } t \text{ to be the compund proposition }p\vee q,\text{then we cat see that}\hspace{1in}\\\begin{aligned}(p\vee q)\wedge r&\equiv t\vee r&&\text{by subsitution of }t\equiv p\vee q\\&\equiv r\wedge (p\vee q) && \text{by Commutative Laws and subsitution } p\vee q \equiv t \\&\equiv (r\wedge p)\vee(r\wedge q)&&\text{by Distributive Laws} \\&\equiv(p ∧ r) ∨ (q ∧ r)&& \text{by Commutative Laws again} \end{aligned}\\\hspace{5in}Q.E.D $$
</aside>
$$ \boxed{(p ∧ q) ∨ r ≡ (p ∨ r) ∧ (q ∨ r)} $$
<aside> 💡 Proof:
$$ \text{similar to last proof:}\hspace{3in}\\\begin{aligned}(p\wedge q)\vee r&\equiv r\vee(p\wedge q) && \text{by Commutative Laws(Consider } p ∧ q \text{ as a whole)} \\&\equiv (r\vee p)\wedge (r\vee q)&&\text{by Distributive Laws} \\&\equiv(p \vee r) \wedge (q \vee r)&& \text{by Commutative Laws again} \end{aligned}\\\hspace{5in}Q.E.D $$
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Logical quantifiers are affected by both the statement and by the universe of discourse. (a) Determine the truth value of each of the following quantified statements, if the universe is all integers:
$$ \begin{aligned}&\textbf{(i)}\exist n(n^2=n)&&\textbf{(ii)}\exist n(n^2=2)\\&\textbf{(iii)}\forall n(n^2\ge n)&&\textbf{(iv)}\forall n \Big((n>5)\rarr(n-1\ge1)\Big)\end{aligned} $$
(b) Which of your answers in (a) would change if the universe was all real numbers instead?
<aside> 💡 (a)
| $\textbf{(i)}\exist n(n^2=n)$ | $\textbf{(ii)}\exist n(n^2=2)$ | $\textbf{(iii)}\forall n(n^2\ge n)$ | $\textbf{(iv)}\forall n \Big((n>5)\rarr(n-1\ge1)\Big)$ |
|---|---|---|---|
| T | F | T | T |
| </aside> |
<aside> 💡 (b) If universe was all real numbers, (ii) would be True and (iii) would be False instead.
| $\textbf{(i)}\exist n(n^2=n)$ | $\textbf{(ii)}\exist n(n^2=2)$ | $\textbf{(iii)}\forall n(n^2\ge n)$ | $\textbf{(iv)}\forall n \Big((n>5)\rarr(n-1\ge1)\Big)$ |
|---|---|---|---|
| T | T | F | T |
| </aside> |