User:Justinlebar/Sandbox

We want to show $\forall x,y\ x\leq y\rightarrow f(x)\leq f(y)\Rightarrow \forall a,b\ f(a\wedge b)\leq f(a)\wedge f(b)$.

Assume the opposite: $\forall x,y\ x\leq y\rightarrow f(x)\leq f(y)\ \mathrm {and} \ \exists a,b\ f(a\wedge b)\leq f(a)\wedge f(b)$.

Now, either $a\leq b$ or $a\nleq b$. In order to arrive at an overall contradiction, we must show that we get a contradiction in either case. Suppose $a\nleq b$. Then we vacuously satisfy $a\leq b\rightarrow f(a)\leq f(b)$. Now we have to show that it's a contradiction that $a\nleq b\ \mathrm {and} \ f(a\wedge b)\leq f(a)\wedge f(b)$ given no information about $f(a)$ or $f(b)$.

Choose f so $f(a)=f(b)=\perp$. Then it must be that $f(a\wedge b)=\perp$; otherwise, it's not true that $f(a\wedge b)\leq f(a)\wedge f(b)$. But I see no reason why $f(a\wedge b)$ should have any specific value.