By definition, a third row ladder escape is a pattern P, for that holds: A+n+P is a template for all n≥0. This equivalent to: A+P and A+1+P are templates.

Proof. That the condition is necessary is obvious. That it is sufficient follows by induction from the following: For all n≥0, If A+n+P and A+(n+1)+P are templates, then A+(n+2)+P is a template, too. It suffices to consider n=0, because one can afterwards substitute P with (n+P) (and this addition is associative :-)). The proposition is true for n=0, because depending on where blue plays, red can reduce A+2+P to either A+P or A+1+P:

If blue plays anywhere except b2, a3 or b3, then red plays b2 and connects immediately. If blue plays b2, then red plays c1 and reduces the position to A+1+P. If blue plays a3 or b3, then red first plays at b2, forcing blue to play b3 or a3. Afterwards red plays d1 and reduces the position to A+P. ∎

In practice, in showing that A+1+P is a template, one may assume that blue drops to the first line, because otherwise red can reduce to A+P. This means:
P is a third row ladder escape if and only if A+P and C+P are templates, where A and C are the patterns

and

.
Perhaps there is an equivalent proposition which requires even fewer amount of work to decide, I don't know.

For the same reason, i.e. that red may raise the ladder at will, if a pattern P is a third row ladder escape, then D+P is a second row ladder escape, where D is

.
That means every third row ladder escape is a second row ladder escape, if we move the beginning of the ladder two cells to the left, and assume that these 6 cells in D are empty. The ladder latter is natural to assume, since we previously expected a third ladder to come from there. For example, let P be the following pattern:

With the proposition from above, one can see that this is a third row ladder escape. Therefore, D+P, that is

is a second row ladder escape. The latter is the same as your last example from the section "2nd row ladder escapes", except with an unnecessary cell on the fourth row.

At the same time,

and

are no second row ladder escapes, so in this case it is really necessary to move the start of the ladder by two cells.

--Wurfmaul (talk) 14:25, 29 May 2016 (UTC)

Two minimaller third row ladder escapes?

It appears to me like two of the presented third row ladder escapes are actually not minimal, since the plusses may be moved two cells to the right to obtain

respectively

.
--Wurfmaul (talk) 15:13, 29 May 2016 (UTC)

This has been fixed.