Schedule algebra

Here is another look at the outcome table. First I complete the left column:

condition on x	on y	replace \([a, b]\) with
\(x > b\)		\([a, b]\)
\(x = b\)		\([a, x-1] [x, y]\)
\(a < x < b\)	\(y < b\)	\([a, x-1] [x, y] [y+1,b]\)
\(a < x < b\)	\(y \ge b\)	\([a, x-1] [x, y]\)
\(a = x\)	\(y < b\)	\([x, y] [y+1, b]\)
\(a = x\)	\(y \ge b\)	\([x, y]\)
\(a > x\)	\(y < a\)	\([a, b]\)
\(a > x\)	\(y = a\)	\([x, y] [y+1, b]\)
\(a > x\)	\(b > y > a\)	\([x, y] [y+1, b]\)
\(a > x\)	\(b \le y\)	\([x, y]\)

Then I reorder by number of elements in the replacement:

condition on x	on y	replace \([a, b]\) with
\(x > b\)		\([a, b]\)
\(a = x\)	\(y \ge b\)	\([x, y]\)
\(a > x\)	\(y < a\)	\([a, b]\)
\(a > x\)	\(b \le y\)	\([x, y]\)
\(x = b\)		\([a, x-1] [x, y]\)
\(a < x < b\)	\(y \ge b\)	\([a, x-1] [x, y]\)
\(a = x\)	\(y < b\)	\([x, y] [y+1, b]\)
\(a > x\)	\(y = a\)	\([x, y] [y+1, b]\)
\(a > x\)	\(b > y > a\)	\([x, y] [y+1, b]\)
\(a < x < b\)	\(y < b\)	\([a, x-1] [x, y] [y+1,b]\)

Then I will do some computations on a piece of paper (not shown here) to derive a simplification.