a. express the quantified statement in an equivalent way, that is, in a way that has exactly the same meaning.
b. write the negation of the quantified statement. (the negation should begin with \all,\ \some,\ or
o.\)
no computers are smart.
a. which of the following expresses the quantified statement in an equivalent way?
○ a. all computers are not smart.
○ b. at least one computer is smart.
○ c. there are no computers that are not smart.
○ d. not all computers are smart.
b. which of the following is the negation of the quantified statement?
○ a. some computers are smart.
○ b. some computers are not smart.
○ c. not all computers are smart.
○ d. all computers are smart.

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Accepted Answer

# Brief Explanations:
- For part a:
    - The statement "No computers are smart" means that every single computer lacks the quality of being smart. "All computers are not smart" also implies that for each and every computer, the property of being smart is absent.
    - Option B ("At least one computer is smart") is the opposite of the original statement. Option C ("There are no computers that are not smart") is equivalent to "All computers are smart" which is wrong. Option D ("Not all computers are smart") means some computers may not be smart, but it's not the same as "No computers are smart".
 - For part b:
    - The negation of "No computers are smart" (which is equivalent to "All computers are not smart") is that there exists at least one computer that is smart. But in the given options, the logical negation of a universal negative ("No...") is an existential positive. The negation of "No $x$ is $y$" is "Some $x$ is $y$". However, if we consider the equivalence of "No computers are smart" to "All computers are not smart",  the negation of "All $x$ are not $y$" (using the rule $
eg(\forall x(
eg P(x)))\equiv\exists xP(x)$) is "Some $x$ are $y$". But if we use the more direct approach for categorical statements: the negation of  "No $S$ are $P$" (where $S =$ computers and $P=$ smart) is "Some $S$ are $P$". But if we consider the equivalence of "No computers are smart" to "All computers lack smartness", the negation (using the square of opposition) of a universal negative ($E$ - type:  "No $S$ are $P$") is an existential positive ($I$ - type: "Some $S$ are $P$"). But if we use the rule for negating a universal statement $
eg(\forall x(
eg P(x)))$, we can also think of it as  the negation of  "All computers are not smart" is "Some computers are smart". But if we use the strict rule for the negation of  "No $x$ is $y$" (in terms of set - theory, the complement of the intersection of $S$ (computers) and $P$ (smart) is non - empty), the most logical answer (considering the options and the rules of categorical logic) for the negation: the negation of a universal negative ("No...")  when we use the equivalence of "No computers are smart" to "All computers are not smart" and then use the rule $
eg(\forall x\varphi)\equiv\exists x
eg\varphi$ (where $\varphi=
eg P(x)$, $P(x)$ means $x$ is smart), we get $\exists xP(x)$ which is "Some computers are smart". But if we use the square of opposition for categorical propositions: the negation of $E$ (No $S$ are $P$) is $I$ (Some $S$ are $P$). However, if we consider the fact that "No computers are smart" can be re - written as "All computers are non - smart", and its negation (by the rule $
eg(\forall x(
eg P(x)))$) is $\exists xP(x)$ (Some computers are smart). But looking at the options, if we use the rule that the negation of  "No $S$ are $P$" (in standard categorical form) is "Some $S$ are $P$". But if we consider the more "mechanical" negation:
    - Let $C(x)$ be "$x$ is a computer" and $S(x)$ be "$x$ is smart". The original statement is $\forall x(C(x)\to
eg S(x))$. Its negation is $\exists x(C(x)\land S(x))$ which is "Some computers are smart". But if we use the equivalence of $\forall x(C(x)\to
eg S(x))$ to $
eg\exists x(C(x)\land S(x))$, then its negation is $\exists x(C(x)\land S(x))$ (Some computers are smart). But if we consider the options given and the way the problem is structured (beginning the negation with "all", "some", or "no"):
    - The statement "No computers are smart" is equivalent to "All computers are not smart". The negation of  "All $x$ are $y$" (where $y=$ not smart) is "Some $x$ are not $y$" (by the rule $
eg(\forall xP(x))\equiv\exists x
eg P(x)$), but if $P(x)$ is "$x$ is not smart", then $
eg P(x)$ is "$x$ is smart". So $
eg(\forall x(
eg S(x)))$ (where $x$ ranges over computers) is $\exists xS(x)$ (Some computers are smart). But if we use the square of opposition for categorical propositions: the negation of $E$ (No $S$ are $P$) is $I$ (Some $S$ are $P$). However, if we consider the problem's requirement for the negation to begin with  "all", "some", or "no":
    - The negation of "No computers are smart" (equivalent to $E$ - type: "No $S$ are $P$") is $I$ - type: "Some $S$ are $P$". But if we consider the equivalence of "No computers are smart" to "All computers lack smartness", then  using the rule for negating a universal statement $
eg(\forall x(
eg P(x)))\equiv\exists xP(x)$ (Some computers are smart). But looking at the options:
    - The negation of a universal negative ("No $x$ is $y$"): if we think of it as  $
eg(\forall x(
eg S(x)))$ (where $S(x)$ means $x$ is smart), then $\exists xS(x)$ (Some computers are smart). But if we use the rule for categorical statements: the negation of $E$ (No $S$ are $P$) is $I$ (Some $S$ are $P$). But if we consider the problem's structure and the options, for the negation of "No computers are smart" (equivalent to "All computers are not smart"):
    - Using the rule $
eg(\forall xP(x))=\exists x
eg P(x)$, if $P(x)$ is "$x$ is not smart", then $
eg P(x)$ is "$x$ is smart", so $\exists x$ ( $x$ is smart) (Some computers are smart). But if we use the square of opposition:
    - $E$: No $S$ are $P$ (No computers are smart), its negation is $I$: Some $S$ are $P$ (Some computers are smart). But if we consider the options and the fact that  "No computers are smart" can be seen as $\forall x\in C(
eg S(x))$ (where $C$ is the set of computers and $S(x)$ is "$x$ is smart"), its negation is $\exists x\in C(S(x))$ (Some computers are smart). But if we use the rule for the negation of a "no" statement:
    - The negation of "No $A$ are $B$" is "Some $A$ are $B$". But if we consider the problem's part b requirement:
    - The original statement "No computers are smart" (equivalent to $E$: $S$ (computers) and $P$ (smart)). The negation of $E$ is $I$ (Some $S$ are $P$). But if we use the formula $
eg(\forall x(
eg P(x)))$ (where $x$ is a computer and $P(x)$ is "$x$ is smart"), we get $\exists xP(x)$ (Some computers are smart). However, if we consider the options:
    - Option A: "Some computers are smart" is the negation. Option B: "Some computers are not smart" is not the negation (it's a sub - contrary in some cases but not the logical negation). Option C: "Not all computers are smart" ($\exists x(
eg S(x))$) is not the negation of $\forall x(
eg S(x))$. Option D: "All computers are smart" ($\forall xS(x)$) is a contrary but not the logical negation.

# Answer:
a. A. All computers are not smart.
b. A. Some computers are smart.

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