0
$\begingroup$

Every natural number is a product of two natural numbers.

$N(x)$ for $x$ is a natural number and $a,b \in\mathbb{N}$

$$\forall x(N(x)\rightarrow \exists a\exists b(N(a)\wedge N(b)\wedge (x=a\cdot b)))$$

Negation:

$$\neg\forall x(N(x)\rightarrow \exists a\exists b(N(a)\wedge N(b)\wedge (x=a\cdot b)))$$

$$\Rightarrow \exists x \neg(N(x)\rightarrow \exists a\exists b(N(a)\wedge N(b)\wedge (x=a\cdot b)))$$

$$\Rightarrow \exists x (N(x)\wedge \neg\exists a\exists b(N(a)\wedge N(b)\wedge (x=a\cdot b)))$$

$$\Rightarrow \exists x (N(x)\wedge \forall a\forall b \neg(N(a)\wedge N(b)\wedge (x=a\cdot b)))$$

$$\Rightarrow \exists x (N(x)\wedge \forall a\forall b (N(a)\rightarrow N(b)\rightarrow (x \neq a\cdot b)))$$

$\endgroup$
2
$\begingroup$

All is good except for the last step, converting $\neg(N(a)\wedge N(b)\wedge (x=a\cdot b)))$. Consider these equivalences: $$ \begin{align} \neg (p_1 \wedge p_2 \dots \wedge p_n \wedge q) &\iff \neg (p_1 \wedge p_2 \dots \wedge p_n) \vee \neg q \quad\text{(by De Morgan)} \\ &\iff (p_1 \wedge p_2 \dots \wedge p_n) \implies \neg q \end{align} $$ So the matrix (inner, unquantified formula) of your result should be $$ (N(a)\wedge N(b)\implies (x\neq a\cdot b)) \text{.} $$

$\endgroup$
0
$\begingroup$

Everything looks good except for changing it to $N(a) \implies N(b)$

$\endgroup$
  • $\begingroup$ Why should it remain $neg(N(a)\wedge N(b)$ ? $\endgroup$ – nikolita Nov 1 '15 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.