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@@ -54,3 +54,39 @@ Quantifiers **inside** the scope of other quantifiers are called **nested quanti
 	- A true statement would be: `Ǝy P(y) -> q ≡ ∀y(P(y) -> q)`
 
 ---
+
+# 1.5 Nested Quantifiers p2
+
+### Converting Sentences to Logical Statements
+
+Ex. Let the domains of x and y be SCU students. Let R(x,y) be "x is related to y," let L(x,y) be "x likes y" and let S(x) be "x sleepwalks."
+- Translate the following sentences into logical statements
+1. There is a student who likes all other students but is related to no others
+	- "There is a student x s.t. for all students y, if y is different from x, then x likes y and x is not related to y"
+	- `Ǝx ∀y[(y!=x) -> L(x,y) ^ ┓R(x,y)]`
+	- **Note**: It's not `Ǝx ∀y[(y!=x) ^ (L(x,y) ^ ┓R(x,y))]`
+		- "There is a student x s.t. all students are not x, x likes y, and x is not related to y"
+		- This proposition will always be false
+	- **Note**: It's not `Ǝx[∀y(y!=x) -> (L(x,y) ^ ┓R(x,y))]`
+2. No one likes a sleepwalker (not even themselves)
+	- "There is no student x for which there is a student who sleepwalks and who x likes"
+	- `┓Ǝx Ǝy[S(y) ^ L(x,y)]`
+		- This can be simplified to `∀x ∀y[┓S(y) ∨ ┓L(x,y)]`
+
+3. `∀x Ǝy[(y!=x) ^ ┓L(y,x)]`
+	- "For every student x, there is a student y s.t. y and x are different and y doesn't like x"
+	- For every SCU student, there is another student SCU student who doesn't like them
+	- Why is this not the same as `∀x Ǝy[(y!=x) -> ┓L(y,x)]`?
+		- Suppose x is liked by everyone and y=x. The statement would be true (doesn't match the sentence)
+
+---
+
+Ex. Let S(x) be "x is a student," A(x,y) be "x has asked y a question," and F(x) be "x is a faculty member." The domain of all variables is people at SCU.
+- Use logical symbols to express: "Some student has never been asked a question by a faculty member"
+	- "There is a person x s.t. x is a student (S(x)), if y is any person who is a faculty member, then y has not asked x a question (F(y) -> ┓A(y,x))"
+	- `Ǝx[S(x) ^ ∀y[F(y) -> ┓A(y,x)]]`
+	- Alternatively, `Ǝx ∀y[S(x) ^ (F(y) -> ┓A(y,x))]`
+	- **Not** `Ǝx ∀y[(S(x) ^ (F(y)) -> ┓A(y,x))]`
+		- True if S(x) or F(y) is false (contradicts the statement)
+
+---