Reducing Boolean expression to an equivalent form

Question

Consider two points from an arbitrary subset of $\mathfrak{R}^n$, namely, points $$p_1 = (x_{11}, x_{12}, \dots, x_{1n})$$ and $$p_2 =(x_{21}, x_{22}, \dots, x_{2n})$$

Let $P_1$ denote the conjunction

$$(x_{11} \le x_{21}) \land (x_{12} \le x_{22}) \land \dots \land (x_{1n} \le x_{2n})$$

and let $P_2$ denote the disjunction

$$(x_{11} < x_{21}) \lor (x_{12} < x_{22}) \lor \dots \lor (x_{1n} < x_{2n})$$

Let $P_3$ denote the following conjunction

$$\lnot(x_{11} > x_{21}) \land \lnot(x_{12} > x_{22}) \land \cdots \land \lnot(x_{1n} > x_{2n})$$

Does $P_1 \land P_2$ imply $P_3 \land P_2$? Or simply $P_3$?

Any input much appreciated. Reference material welcome.

Answer 1

$P_3$ is equivalent to $P_1$ (because individually each $x_{1k} \le x_{2k}$ is equivalent to $\lnot(x_{1k}>x_{2k})$, for $k$ between $1$ and $n$).

So $P_1 \land P_2$ does imply $P_3 \land P_2$ (and in fact the two are equivalent).

Answer 2

Given points $\mathrm p, \mathrm q \in \mathbb R^n$, let

$$\Phi_1 := \bigwedge_{i=1}^n (p_i \leq q_i) \equiv \color{blue}{\mathrm p \leq \mathrm q}$$

$$\Phi_2 := \bigvee_{i=1}^n (p_i < q_i) \equiv \neg \bigwedge_{i=1}^n \neg (p_i < q_i) \equiv \neg \bigwedge_{i=1}^n (p_i \geq q_i) \equiv \color{blue}{\neg (\mathrm p \geq \mathrm q)}$$

$$\Phi_3 := \bigwedge_{i=1}^n \neg (p_i > q_i) \equiv \bigwedge_{i=1}^n (p_i \leq q_i) \equiv \color{blue}{\mathrm p \leq \mathrm q}$$

Note that $\Phi_1 \equiv \Phi_3$. Note also that $\neg (\mathrm p \geq \mathrm q) \equiv \mathrm p < \mathrm q$ only when $n=1$.