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.