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 (is this actually true? I think it is.). 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.

Alternate approach:

Statement 1: $\forall x,y\ x\leq y\rightarrow f(x)\leq f(y)$

Statement 2: $\forall a,b\ f(a\wedge b)\leq f(a)\wedge f(b)$

Proof of $1\Rightarrow 2$:

Consider arbitrary $x$ and $y$.

By the properties of $\wedge$, $a\wedge b\leq a$ and $a\wedge b\leq b$.

Thus $x\wedge y\leq x$ and $x\wedge y\leq y$.

Applying Statement 1, $f(x\wedge y)\leq f(x)$ and $f(x\wedge y)\leq f(y)$.

This means that $f(x\wedge y)$ is a lower bound for $f(x)$ and $f(y)$.

We defined $f(x)\wedge f(y)$ as the greatest lower bound of $f(x)$ and $f(y)$, so $f(x\wedge y)\leq f(x)\wedge f(y)$.

Proof of $2\Rightarrow 1$:

Let $x$, $y$ be such that $x\leq y$.

By the definition of the $\leq$ relation, $x\wedge y=x$.

By statement 2, $f(x\wedge y)\leq f(x)\wedge f(y)$.

Substituting, $f(x)\leq f(x)\wedge f(y)$.

By the properties of $\wedge$, $a\wedge b\leq a$ and $a\wedge b\leq b$.

Thus $f(x)\leq f(x)\wedge f(y)\leq f(y)$.