From 01aab0788a4451cd82d0ed56ce2b21cb825e6e1f Mon Sep 17 00:00:00 2001 From: Louie S Date: Wed, 8 Apr 2020 09:58:58 -0700 Subject: Post-class 04/08 --- 1.5.md | 36 ++++++++++++++++++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/1.5.md b/1.5.md index 4f8108c..3bdb0ab 100644 --- a/1.5.md +++ b/1.5.md @@ -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) + +--- -- cgit