Question: I'm simulating a auction game with sealed single bid, where each of the $n$ players has winning probability $p\_i,i=1,...,n$, and their bids $b\_i$ have to be calculated to meet the $p\_i$. Supposing that the value $v\_i$ of each player is uniformly distributed in the interval $[0,1)$ and $b\_i$ is a function of $v\_i$ (as usual), how could I calculate their bids to match the $p\_i$?

Answer: 
DeMorgan's laws:

1. $\sim (x \vee y) = (\sim x ) \wedge (\sim y)$
2. $\sim (x \wedge y) = (\sim x ) \vee (\sim y)$

In some notations where the overline indicates the complement, the rule is "break the line, change the sign":

1. $\overline{x+y} = \overline{x} \ \overline{y}$
2. $\overline{xy} = \overline{x} + \overline{y}$

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By only manipulating the right hand side, we have
$$( \sim q \ \lor \sim p) \wedge (\sim r \ \lor \sim p) \\ 
= [\sim (q\wedge p)] \wedge [\sim (r \land p)] \\ 
= \sim [(q\wedge p) \lor (r \land p)] \\
= \sim [(p \land q) \lor (p \land r)] \\
= \sim [p \land (q \lor r)] $$

which is contradictory (false in all possible worlds) to the original left hand side.
I'll show you why the left-hand side and left hand side of the claim are contradictory: $$\begin{align} \left[p \wedge (q \lor r)\right] &\equiv \left[( \lnot q \lor \lnot p) \wedge (\lnot r \ \ \lor \lnot p)\right]\tag{1}\\ \\ \iff (p\land q )\lor (p \land r) &\equiv \lnot(q \land p) \land \lnot(r \land p)\tag{2}\\ \\ \iff (p\land q )\lor (p \land r) &\equiv \lnot[(p\land q) \lor (p\land r)]\tag{3}\end{align}$$

$(1 \to 2)$: Distributive law (left), DeMorgan's, twice (right)

$(2\to 3)$ DeMorgan's Law, applied on the left.

This is clearly a self-contradictory proposition, since the right-hand side is the negation of the left-hand side. This means the claim is false in any/all possible worlds, i.e., it is a contradiction (always false, regardless of the truth values of $p, q, r)$.