From f5b740bca69154e8f6435078cad1a7c092243790 Mon Sep 17 00:00:00 2001
From: Louie S <lshprung@yahoo.com>
Date: Fri, 3 Apr 2020 10:10:59 -0700
Subject: Post-class 04/03

---
 1.3.md | 30 ++++++++++++++++++++++
 1.4.md | 90 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 120 insertions(+)
 create mode 100644 1.4.md

diff --git a/1.3.md b/1.3.md
index a46f0e4..86af4e4 100644
--- a/1.3.md
+++ b/1.3.md
@@ -99,3 +99,33 @@ Truth Table:
 	- with a truth table
 	- by (3)
 3. Use table of Common Equivalences (see link above)
+
+---
+
+# Using the Table of Common Equivalences
+
+- Ex. Show `(p^q)->(p->q)` is a tautology without a truth table
+	- We are goint to solve this using a proof
+	- **Goal**: `(p^q)->(p->q)≡T`
+
+|Proof: (p^q)->(p->q)|Reason|
+|--------------------|------|
+|≡(p^q)->(┓p∨q)      |Implication|
+|≡┓(p^q)∨(┓p∨q)      |Implication|
+|≡(┓p∨┓q)∨(┓p∨q)     |De Morgan|
+|≡┓p∨(┓q∨(┓p∨q))     |Associative|
+|≡┓p∨(┓q∨(q∨┓p))     |Commutative|
+|≡┓p∨((┓q∨q)∨┓p))    |Associative|
+|≡┓p∨(T∨┓p))         |Negation|
+|≡┓p∨T               |Domination|
+|≡T                  |Domination|
+
+- Why use this method?
+	- This method is useful for compound propositions with many propositions
+
+- `n` propositions -> truth table has 2^n rows
+	- e.g 5 propositions -> truth table has 32 rows (!)
+
+---
+
+[1.4 ->](1.4.md)
diff --git a/1.4.md b/1.4.md
new file mode 100644
index 0000000..db27f54
--- /dev/null
+++ b/1.4.md
@@ -0,0 +1,90 @@
+[\<- 1.3](1.3.md)
+
+---
+
+# 1.4 Predicates and Quantifiers p1
+
+## Propositional Functions
+
+- Ex. Note "w < 5" and "x^2 + y^2 = z^2" are not propositions since a proposition is either true or false, but not both.
+	- The problem is they could be true or false depending on what the variables are
+
+- Let P(w) be "w < 5"
+	- Once we specify a value for "w", this becomes a proposition
+	- E.g. P(3) -> 3<5 -> `P(3)≡T`
+	- E.g. P(6) -> 6<5 -> `P(6)≡F`
+	- We call P(w) a **propositional function**
+
+- Ex. Let R(x,y,z) be the propositional function "x^2 + y^2 = z^2". Evaluate the truth values of R(1,2,3) and R(3,4,5)
+	- R(1,2,3) -> 1 + 4 = 9 -> `R(1,2,3)≡F`
+	- R(3,4,5) -> 9 + 16 = 25 -> `R(3,4,5)≡T`
+
+---
+
+## Quantifiers
+
+- The proposition "for all (for every) x in the domain, P(x) is true" is denoted `∀x P(x)`
+	- `∀x P(x)` is read "for all x P(x) is true
+	- `∀` = "for all" and is called the universal quantifier
+
+- Ex. Let P(x) be "x^2 >= 0" and Q(x) be "x^2 > 0". If the domain is all reals, evalueate the truth values of the propositions
+	- (a) `∀x P(x)`
+		- `∀x P(x) ≡ T`
+	- (b) `∀x Q(x)`
+		- `∀x Q(x) ≡ F` since "x^2 >= 0" is *not* true when x=0.
+		- Note: If there is even one value of x where Q(x) is false, then `∀x Q(x) ≡ F`
+
+- Ex. Let L(x) be "x > 8" with domain {10, 11, 12}. Then,
+	- `∀x L(x) ≡ L(10) ^ L(11) ^ L(12) ≡ T`
+	- For a **finite domain**, we can rewrite `∀x L(x)` as a **conjunction**
+
+---
+
+- The proposition "there exists (there is) an x in the domain such that P(x) is true" is denoted by `Ǝx P(x)`
+	- `Ǝ` is read "there exists" and is called the existential quantifier
+
+- To review:
+	- `Ǝx` means "there's at least one x in the domain..."
+	- `∀x` means "for all x in the domain, ..."
+
+- Ex. Let M(x) be "2x + 3 = 29". Evaluate the truth value of `Ǝx M(x)` if the domain is
+	- (a) {10,11,12}
+		- `Ǝx M(x) ≡ M(10) ∨ M(11) ∨ M(12)`
+			- `M(10) ≡ F`
+			- `M(11) ≡ F`
+			- `M(12) ≡ F`
+			- Therefore, `Ǝx M(x) ≡ F`
+	- (b) Integers
+		- `Ǝx M(x) ≡ T` since x=13 makes M(x) true
+	- For a **finite** domain, we can write `Ǝx M(x)` as **disjunction** (or statement)
+	
+---
+
+# 1.4 Predicates and Quantifiers p2
+
+- The propositional function "there exists a **unique** x in the domain such that P(x) is true" is denoted by `Ǝ!x P(x)`
+	- The "!" means "unique"
+
+- Ex. Let the domain of x be the reals. Let T(x) be "x^3 - 4 = 0" and S(x) be "x^2 = 4x". Evaluate the truth value of
+	- (a) `Ǝ!x T(x)`
+		- `Ǝ!x T(x) ≡ T` since x = 4^(1/3) is the **only** value that works
+	- (b) `Ǝ!x S(x)`
+		- How to apporach this problem:
+			- **Incorrect** dividing both sides by x (you get x=4) causes you to lose solutions
+			- **Correct** move everything to one side and factor (you get x(x-4)=0)
+		- `Ǝ!x S(x) ≡ F` since both x=0 and x=4 work (there is more than one x that works)
+
+### Two ways to turn Propositional Function, R(x), into a Proposition
+
+1. Assign a value to x
+2. Put a quantifier in front of it
+
+### Precedence
+
+1. `∀`, `Ǝ`, `Ǝ!`
+2. `┓`
+3. `^`, `∨`, `⊕`
+4. `->`, `<->`
+
+- Ex. `∀x P(x) -> q`
+	- means `(∀x P(x)) -> q`
-- 
cgit