I'll explain what I mean by orthogonal, which is probably a poor choice of words on my part.
Given two Turing machines $\lambda $ and $\tau$,and two inputs $i$ and $j$. lets say $\tau(i) \preceq \lambda(j)$, if there is a proof $P$, that states that $\tau(i)$ halts if $\lambda(j)$ halts. As an example of where we might have $\tau(i) \preceq \lambda(i)$. For instance, $\lambda$ may simply perform the computation $1+1$, and then run $\tau(i)$.
What I mean by orthogonal is noncomparable under this order. That is does there exist a pair of turing machines $\lambda$ and $\tau$ and inputs $i$ and $j$, where neither $\lambda(j) \preceq \tau(i)$ nor $\tau(i) \preceq \lambda(j)$
I believe I can show that for a fixed non-halting $\tau(i)$, we cannot have $\tau(i) \preceq \lambda(j)$ for all $\lambda(i)$. Since we could then use this to create a machine to solve the halting problem, by enumerating proofs until we find a proof that $\tau(i) \preceq \lambda(j)$. The existence of a proof then gives us that $\lambda(j)$ does not halt as $\tau(i)$ does not.
However, this result is far weaker than my goal.