Find all the estimations that show that the following is false. $(x \land y)\lor (x\land z)\lor(y\land z)\lor(u\land v)\lor (u\land w)\lor(v\land w)\lor(\neg x\land\neg u)$. I know how to solve this by using truth tables however, I need 64 lines.
How can I simplify it, or use another way to solve it?
Since this is a disjunction, for it to be false you need each of the propositions in the disjunction to be false.
Each of the propositions is a conjunction, so in order to be false, you just need one of the two to be false.
The first three and fourth through sixth are symmetric, so the analysis for those for $x,y,z$ will be the same as for $u,v,w$ before we get to the last one.
For $x\wedge y$, $y\wedge z$, and $x\wedge z$ to all be false, you need at least two of $x,y,z$ to be false; at most one can be true. Likewise for $u$, $v$, and $w$.
Now, in the last one you have $\neg x\wedge\neg u$. For this to be false, you need either $x$ to be true (which forces $y$ and $z$ to be false), or $u$ to be true (which forces $v$ and $w$ to be false), or both.
So, when is this full proposition false?
When $x$ is true and $u$ is false, in which case $y$ and $z$ are false, and at most one of $v$ and $w$ is true (three possibilities total).
When $x$ is false and $u$ is true, in which case $v$ and $w$ are both false and at most one of $y$ and $z$ is true (three possibilities total).
When both $x$ and $u$ are both true, in which case $y$, $z$, $v$, and $w$ are all false.
That gives 3 assignments for case 1, three assignments for case 2, and one assignment for case 3, giving you all seven situations in which the proposition is false.
