2

I have series of statement that can logically be true($1$) or false($0$). For example if I have just one statement i.e

  1. Statement $1$ is false.

Statement $1$ says that itself is false which is contradictory, so I can conclude that the statement is, in conclusion, contradictory. But if I have $2$ statements such that

  1. Statement $2$ is false
  2. Statement $1$ is false

If statement $1$ is true, then statement $2$ must be false which implies that statement $1$ is true. If statement $2$ is true then statement $1$ must be false which in turn implies that statement $2$ is true. We see that there are no contradictions between these two statements.

But of course the two examples are simple cases. Assuming that I have the following $10$ statements (I can have more, but for example)

  1. Statement $5$ is true.
  2. Statement $7$ is false.
  3. Statement $4$ is true.
  4. Statement $5$ is false.
  5. Statement $3$ is true.
  6. Statement $7$ is false.
  7. Statement $1$ is false.
  8. Statement $8$ is true.
  9. Statement $10$ is false.
  10. Statement $1$ is false.

What procedures do I have to take to or simply how do I successfully determine the statements that correlate and those that contradicts?

1 Answers1

5

Let $a_1, a_2, \ldots, a_n$ be variables, each equal to $1$ (true) or $-1$ (false). Then your statements become equations.

If Statement $i$ says Statement $j$ is false, that means $a_i=-a_j$.

If Statement $i$ says Statement $j$ is true, that means $a_i=a_j$.

In your example, $a_1=a_5, a_2=-a_7, a_3=a_4, a_4=-a_5, a_5=a_3, \ldots$. Already you have a contradiction among $a_3, a_4, a_5$.

In general, if there is one solution, there will be at lesat two (everyone swaps true, false). So pick $a_1$ to be true, and trace along the equations. If $a_1$ is true, then $a_5$ is true, then $a_3$ is true, then $a_4$ is true, then $a_5$ is false (contradiction). If you got to a cycle without a contradiction, look at the unused equations. If any of them connect to what you've built so far, then continue; otherwise you have two disjoint components (e.g. $a_1=a_2, a_3=a_4=-a_5$).

vadim123
  • 82,796