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