How do I negate the following statement?
Also please help me with this exercise:
First question : $$\neg(c): (\exists x,y \in \Bbb R)(\exists z \in \Bbb Z)(\forall a \in \Bbb R) \quad a \geq \frac{x+y}{2} \vee a > z $$
Second question:
Let $x \in U$ where $U$ is the universal set indicated above. Then:
$$x \in (A\backslash B) \backslash C \iff x \in A \wedge x \notin B \wedge x \notin C $$ Let us call $(1)$ the right proposition of this equivalence.
$$x \in A\backslash(B \backslash C) \iff x \in A \wedge x \notin B \backslash C$$ Let us call $(2)$ the right proposition of this equivalence.
if $x \in A$ and $x \notin B $ and $x \in C$ then $(2)$ holds but not $(1)$
This allow us to construct a counter-example : let $A=C=\{1\}$ and $B=\{2\}$
Then $(A\backslash B) \backslash C= \emptyset$ and $A \backslash (B \backslash C)=\{1\}$
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First question : $$\neg(c): (\exists x,y \in \Bbb R)(\exists z \in \Bbb Z)(\forall a \in \Bbb R) \quad a \geq \frac{x+y}{2} \vee a > z $$
Second question:
Let $x \in U$ where $U$ is the universal set indicated above. Then:
$$x \in (A\backslash B) \backslash C \iff x \in A \wedge x \notin B \wedge x \notin C $$ Let us call $(1)$ the right proposition of this equivalence.
$$x \in A\backslash(B \backslash C) \iff x \in A \wedge x \notin B \backslash C$$ Let us call $(2)$ the right proposition of this equivalence.
if $x \in A$ and $x \notin B $ and $x \in C$ then $(2)$ holds but not $(1)$
This allow us to construct a counter-example : let $A=C=\{1\}$ and $B=\{2\}$
Then $(A\backslash B) \backslash C= \emptyset$ and $A \backslash (B \backslash C)=\{1\}$