User:Dfeuer/Properties of Heyting Algebras
Let $(L, \wedge, \vee, \preceq)$ be a Heyting algebra.
Let $x, y, z \in L$.
Then:
Lemma
$x \to y = \top \iff x \preceq y$
Proof
Suppose $x \preceq y$.
Then $x \land \top = x \preceq y$, so $x \to y = \top$.
Suppose $x \to y = \top$.
Then $x \land \top \preceq y$.
But $x \land \top = x$, so $x \preceq y$.
Thm 1
If $x \to y = top$ and $y \to x = \top$ then $x = y$.
Proof
We have $x \land \top \preceq y$ and $y \land \top \preceq x$.
Thus $x \preceq y$ and $y \preceq x$, so $x = y$.
Thm 2
If $\top \to x = \top$ then $x = \top$
Proof
We have $\top \land \top \preceq x$, so $\top \preceq x$, so $x = \top$.
Thm 3
$x \to (y \to x) = \top$
Proof
We will show that $x \preceq (y \to x)$. The theorem will then follow from the lemma.
We have that $y \to x$ is the greatest $z$ such that $y \land z \preceq x$.
But $y \land x \preceq x$ by the definition of meet, so $x \preceq z$.
That is, $x \preceq (y \to x)$.
Thm 4
$(x \to (y \to z)) \to ((x \to y) \to (x \to z)) = \top$
Proof
By the lemma, it is sufficient to show that:
- $x \to (y \to z) \preceq (x \to y) \to (x \to z)$
By the definition of relative pseudocomplement, this is equivalent to showing that:
- $(x \to (y \to z)) \land (x \to y) \preceq x \to z$
Let $m = (x \to (y \to z)) \land (x \to y)$.
By the definition of meet, $m \preceq x \to (y \to z)$ and $m \preceq x \to y$.
Then $m \land x \preceq y \to z$ and $m \land x \preceq y$.
Thus $m \land x \land y \preceq z$ and $m \land x \preceq y$.
Since $m \land x \preceq y$, $m \land x \land y = m \land x$.
Thus $m \land x \preceq z$.
By the definition of pseudocomplement, $m \preceq x \to z$.