What is the opposite of this condition? Condition: $(A=1) \land (C>1) \land (B<6)$
Opposite Condition: $(A\ne 1) \lor (C\le 1) \lor (B\ge 6)$
Is that true?  
 A: It's a simple example of De Morgan laws 
A: You are right:
\begin{align*}
&&\lnot((A=1) \land (C>1) \land (B<6))\\
&\Longleftrightarrow&\lnot(A=1) \lor \lnot(C>1) \lor \lnot(B<6) \\
&\Longleftrightarrow&(A \ne 1) \lor (C\le 1) \lor (B \ge 6)
\end{align*}
A: Let's do it step-by-step. For simplification, we assume the standard ordering of the Reals. If we want to really be rigorous, i.e., for a  general ordering relation, then $\nless$ is just the negation of $<$:
Let's use $\lnot$ for negation, and let's group the sentences in pairs; I think it will be easier/clearer:
Condition: $[(A=1) \land (C>1)] \land[ (B<6)]$ , so that we have two sentences; $P:=[(A=1)\land 
(C>1)]$ ; $Q:=[B<6]$ . Set the internal sentences to be $A,C $ respectively.
Opposite Condition for $P$: $\lnot(A \land C)$ is: $\lnot A \lor\lnot C$. Negation of conjunction distributes
as the disjunction of negations.
Let's apply $\lnot$ to $P$ first:
$\lnot P=\lnot [(A=1)\land(C>1)]=\lnot (A=1) \lor\lnot (C>1)\equiv(A\neq 1)\lor (C\nge 1)$.
Now, let's do $\lnot Q$. We have $\lnot[B<6]\equiv[B\nless6]$.
Putting it all together, we get: $(A \neq 1) \lor(C\leq 1)\lor(B\nless 6)$.
