- $A = \{1,2\}$
- $B = \{2,3\}$
- $R \subseteq A \times B$
- $R = \{(1,2), (1,3), (2,3)\}$
I want to prove $R$ is not a function. That is, I want to show: $$ \lnot \forall x(x \in A \to \exists y(y \in B \land (x,y) \in R \land \forall z((z \in B \land (x,z) \in R) \to y = z))) $$
and using logical equivalences I arrive at: $$ \exists x(x \in A \land \forall y(y \in B \land (x,y) \in R \to \exists z((z \in B \land (x,z) \in R) \land y \neq z))) \tag{1} $$
What I do not understand: If I find an $x \in A$ such that $(x,y), (x,z) \in R$ and $y \neq z$, isn't it enough to conclude $R$ is not a function ?
I think this would be the symbolization: $$ \exists x(x \in A \land \exists y(y \in B \land (x,y) \in R \land \exists z((z \in B \land (x,z) \in R) \land y \neq z))) \tag{2} $$
Perhaps, I am reading in an incorrect manner the Negation of Uniqueness Quantifier. But, I cannot see what would be the difference with my last symbolization and also cannot think a case where my last symbolization fails to represent that $R$ is not a function. I will appreciate any insight.
You're last statement assumes that $\operatorname{dom}(R) = A$. Yes, establishing that would prove that $R$ isn't a function. The broader statement with the $\forall y$ quantifier allows for the possibility that $\operatorname{dom}(R) \subsetneq A$: if $x\in A$ but $x \notin \operatorname{dom}(R)$, then that $x$ satisfies the statement trivially, as there are no $y\in B$ for which $(x, y)\in R$.