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@@ -88,3 +88,55 @@ - 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) |