0 0 votes Express the statement "Everyone has exactly one best friend" as a logical expression involving predicates,quantifiers with a domain consisting of all people,and logical connectives without using uniqueness quantifier. I am confused pleased explain it Mathematical Logic propositional-logic + – One 4.1k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 5 5 votes Everyone has exactly one best friend. Let me define few predicates before proceeding to the answer : $F(x,y)$ = $y$ is a Best friend of $x$. And let the domain be All People in the world. (Assuming that No one is a best friend of himself/herself. (i.e. $F(x,x)$ is False) ) $\forall x (\exists y F(x,y) \wedge \forall z ((z \neq y) \rightarrow \sim F(x,z)))$ Interpretation : For every person $x$ there is some person $y$ who is best friend of $x$ And for any(every) person $z$, if $z$ is not same person as $y$ then $z$ is Not a best friend of $x$. Refer here for clarity about Uniqueness Quantifier/Quantification : https://gateoverflow.in/219473/kenneth-rosen-ch-1-ex-1-5-qn-52?show=219480#a219480 Deepak Poonia answered Jun 4, 2018 • selected Jun 5, 2018 by srestha Deepak Poonia comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Instead of including a new variable z we can also express the statement as $\forall x\exists !yB(x,y)$ where $\exists! $ is the "uniqueness quantifier". rohithk answered Dec 28, 2019 rohithk comment Share Follow 0 reply Please log in or register to add a comment.
–1 –1 vote people =p(x) best friend=b(y) ∀x(p(x)⟶∃y(b(y))) srestha answered May 14, 2016 srestha comment Share Follow See 1 comment 1 1 comment reply Deepak Poonia commented Jun 4, 2018 reply Follow flag If $b(x,y)$ means $y$ is a best friend of $x$ and If the domain is taken to be All the people in the world then the expression ∀x(p(x)⟶∃y(b(x,y))) For every person $x$, there is at least one person who is best friend of $x$. It does not say "One and Only One". 0 0 replyShare Please log in or register to add a comment.