0
$\begingroup$

Let $X$ be a set. Let $0$ be an element of $X$.

For a function $P$ defined on tuples of $n$ elements of the set $X$ we know (for every tuples $f$ and $g$ each having $n$ elements) $\forall i \in n : ( f_i \neq 0 \wedge g_i \neq 0 ) \wedge P f = P g \Rightarrow f = g.$

Let $X$ be also a poset with least element $0$.

Under which additional conditions can we prove: $\forall i \in n : ( f_i \neq 0 \wedge g_i \neq 0 ) \wedge P f \le P g \Rightarrow \forall i\in n: f_i \le g_i?$

It is a practical task to prove it, don't be afraid to assume additional conditions. Maybe we should require that $X$ is a (semi)lattice?

  • 0
    Ugh, it seems that there are no simple condition for this (and my question has no answer). Consider for example the case when the orders are dual (reverse) to the formula above. Then the formula above, which I want, does not hold, but it seems to not contradict to any essential condition we may want to specify.2012-07-07

1 Answers 1

0

There is no answer to this question. Consider for example when a reverse of the above formula holds: $\forall i \in n : ( f_i \neq 0 \wedge g_i \neq 0 ) \wedge P f \le P g \Rightarrow \forall i\in n: f_i \ge g_i.$

This case can't be distinguished from what is wanted in the question, without explicitly specifying a partial order for the image of $P$.