Find fitlers in the Boolean algebra $B = {1, 2, 5, 7, 10, 14, 35, 70}$ under the operations $+, \cdot, '$ defined by $$x + y = lcm(x, y)$$ $$x \cdot y = gcd(x, y)$$ $$x' = \frac{70}{x}.$$ I know that $1$ is a unit element and $70$ is a zero element. If $F$ is a filter, then
$(1)$ $1\in F,$
$(2)$ if $x\in F$ and $x\leqslant y,$ then $y\in F,$
$(3)$ if $x,y\in F,$ then $x+y\in F.$
Also, $x\leqslant y$ iff $x+y=x.$
I think filters are $F_1=\{1\},$ $F_2=\{1,2\},$ $F_3=\{1,5\},$ $F_4=\{1,7\},$ $F_5=\{1,2,5,10\},$ $F_6=\{1,2,7,14\},$ $F_7=\{1,5,7,35\},$ $F_8=\{1,2,5,7,10,14,35,70\}.$
Am I right?
The order on $B$ is divisibility. So if $1 \in F$ then all members of $B$ are in $F$. I woould rather all $1$ the unique minimum "$0$" of the BA and $70$ the maximum "$1$", but that might be confusing in this context..
We at least get the fixed filters so sets like $F = \{2,10,14,70\}$ (eveerything above $2$) etc. So your list is definitely not correct.