Note: Book of Proof

2. Logic

DeMorgan's Laws:

$\neg (P \land Q) = (\neg P) \lor (\neg Q)$

$\neg (P \lor Q) = (\neg P) \land (\neg Q)$

$(P \land Q) \implies R$
$P(x) \land \neg P(y)$
$P \implies \neg Q$
$\forall \text{prime } p, \exists \text{prime } q, q > p$
$\forall \epsilon \in \mathbb{R}, \epsilon > 0, \exists \delta \in \mathbb{R}, \delta > 0, (|x - a| < \delta) \implies (|f(x) - f(a)| < \epsilon)$
$\forall \epsilon \in \mathbb{R}, \epsilon > 0, \exists M \in \mathbb{R}, M > 0, (x > M) \implies (|f(x) - b| < \epsilon)$
$\exists a \in \mathbb{R}, \forall x \in \mathbb{R}, a + x = x$
$\neg P$
$(x \in \mathbb{Q} \land x \neq 0) \implies \tan(x) \notin \mathbb{Q}$
$(\sin(x) < 0) \implies \neg (0 \le x \le \pi)$

$x$ is not positive or $y$ is positive.
$x$ is prime, and $\sqrt{x}$ is rational.
There exists a prime number $p$ , for every prime number $q$ , such that $q \le p$ .
There exists a positive number $\epsilon$ , for every positive number $M$ , such that $(x > M)$ and $(|f(x) - b| \ge \epsilon)$ .
$\forall a \in \mathbb{R}, \exists x \in \mathbb{R}, a + x \neq x$
$P: x \in \mathbb{Q}$
$Q: x \neq 0$
$R: \tan(x) \in \mathbb{Q}$
$S: (P \land Q) \implies \neg R$
$\neg S: (P \land Q) \land R$
There exists a number $x$ , such that $(\sin(x) < 0) \land (0 \le x \le \pi)$ .
There exists a function $f$ , where $f$ is a polynomial and its degree is greater than 2, and $f'$ is constant.

(a) $6 \cdot 6 \cdot 6 \cdot 6 = 1296$
(b) $6 \cdot 6 \cdot 6 = 216$
(c) $5 \cdot 6 \cdot 6 \cdot 6 = 1080$
$26 \cdot 26 \cdot 26 = 17576$
(a) $6 \cdot 6 \cdot 6 = 216$
(b) $6 \cdot 5 \cdot 4 = 120$
(c) $6 \cdot 5 = 30$
(d) $6 \cdot 6 = 36$
$2 \cdot 6 = 12$
(a) $2^8 = 256$
(b) $2^7 = 128$