Post-class 04/06

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 - Ex. `∀x P(x) -> q`
 	- means `(∀x P(x)) -> q`
+
+---
+
+## Scope
+
+- `T(x) -> r` is a propositional function
+	- T(x) and r are each propositions for which we do not know the truth values
+- `(∀x T(x)) -> r` is a proposition
+	- **scope** of ∀x is T(x) (clarified by parenthesis)
+- `∀x (T(x) -> r)` is a proposition
+	- **scope** of ∀x is `T(x)->r`
+
+Takeaway: If there's a variable not in the scope of a quantifier, it's**not** a proposition
+
+### Convert English into Logical Statements
+
+- Ex. What is the negation of "all dogs bark"? Write in words and logical symbols (2 ways)
+	- Let B(x) be "x barks" and the domain of x is dogs
+	
+|Words|Symbols|
+|-----|-------|
+|All dogs bark|∀x B(x)|
+|Not all dogs bark|┓∀x B(x)|
+|There is at least one dog who doesn't bark|Ǝx ┓B(x)|
+
+Takeaway:
+- `┓∀x B(x) ≡ Ǝx ┓B(x)`
+- `Ǝx ┓B(x) ≡ ∀x ┓B(x)`
+	- "It's not the case that there is a dog that barks" means the same as "No dogs bark"
+
+---
+
+- Ex. Convert the following sentences into logical statements
+	1. "There is a woman with a PhD and an MD"
+		- Let P(x) be "x has a PhD" and the domain of x is women
+		- Let M(x) be "x has an MD" and the domain of x is women
+		- `Ǝx (P(x) ^ M(x))`
+	2. "There is exactly one woman with a PhD and MD"
+		- Use the setup from the previous sentence
+		- `Ǝ!x (P(x) ^ M(x))`
+	3. "No one has seen an alien"
+		- Think about it: "No one" is the opposite of "at least one"
+		- Let A(x) be "x has seen an alien" where the domain of x is people
+		- `┓Ǝx A(x)` simplified to `∀x ┓A(x)`
+	4. "The campus is quiet when everyone is home"
+		- Let H(x) be "x is home" on the domain of x is people
+		- Let Q be "The campus is quiet"
+		- `(∀x H(x)) -> Q`
+
+---
+
+[1.5 ->](1.5.md)