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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)
diff --git a/1.5.md b/1.5.md
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+[\<- 1.4](1.4.md)
+
+---
+
+# 1.5 Nested Quantifiers p1
+
+- Let the domain of x and y be the reals
+
+|Logical Statement|Words|
+|-----------------|-----|
+|∀x(∀y(xy=xy))    |"For all x∈ℝ (is an element of real numbers) and all y∈ℝ, xy = yx"|
+|∀y(∀x(xy=xy))    |"For all y∈ℝ (is an element of real numbers) and all x∈ℝ, xy = yx"|
+
+- Both columns are true
+- `∀x(∀y(xy=xy)) ≡ ∀y(∀x(xy=xy))`
+
+Quantifiers **inside** the scope of other quantifiers are called **nested quantifiers**
+
+- Note: `∀s ∀t P(s,t) ≡ ∀t ∀s P(s,t)
+
+---
+
+- Ex. Let the domain of x and y be **positive reals**. Convert the logical statement to words and determine its truth value.
+	1. `∀x Ǝy (x=y^2)`
+		- "For all x>0, there exists y>0 s.t. (such that) x=y^2."
+		- This statement is **True** since for any x, we would pick y=sqrt(x)
+			- e.g. if x=4, then y=2
+			- e.g. if x=5.72, then y=sqrt(5.72)
+	2. `Ǝy ∀x (x=y^2)`
+		- "There exists a number y>0 s.t. for all x>0, x=y^2."
+		- This statement is **False** since once y is chosen, y^2 can only be one value. Therefore, only one value of x would satisfy x=y^2 (and not all x).
+		- Note: `∀x Ǝy P(x,y)` is **not** equivalent to `Ǝy ∀x P(x,y)`
+
+- Ex. Let the domain of x and y be integers. Convert the logical statement to words and determine its truth value.
+	1. `Ǝx Ǝy (x^2 + y^2 = 3)`
+		- In words: "There is an integer x and there is an integer y s.t. x^2 + y^2 = 3"
+	2. `Ǝy Ǝx (x^2 + y^2 = 3)`
+		- In words: "There is an integer y and there is an integer x s.t. x^2 + y^2 = 3"
+	- Both of these examples have the same meaning
+		- `Ǝx Ǝy P(x,y) ≡ Ǝy Ǝx P(x,y)`
+	- These statements are **False** since no **integer** values of x and y can be plugged in to create a true statement
+
+---
+
+**Careful**: `∀x(Ǝy P(y) -> ...)` is **not** equivalent to `∀x Ǝy(P(y) -> ...)`
+- Be careful when moving quantifiers to make sure the meaning is the same
+
+- Ex. Let P(Y) be "student y turns in their homework". Domain of y is students in our class. Let q be "Luvreet is happy"
+	- `Ǝy P(y) -> q`
+		- "If there is a student y who turns in their homework, then Luvreet is happy"
+	- `Ǝy(P(y) -> q)`
+		- "There is a student y s.t. if y turns in their homework, then Luvreet is happy"
+	- These propositional functions are different the first one is basically saying, "Luvreet is happy when **any** student turns in their homework" but the second is basically saying, "Luvreet is happy when y turns in their homework"
+	- A true statement would be: `Ǝy P(y) -> q ≡ ∀y(P(y) -> q)`
+
+---