Although replacing MOV with INV or NEG is faster for that single instruction, what is the impact for the overall sequence? Am I just getting wrapped up in all this for no real gains? Time to double check.

I need some shorthand for the compound jumps, so here are the cycle timings for each possible exit from a compund jump:
jlt: 4+(8..10) -> 14 = 14
jeq: 4+(8..10) -> 12+14 = 26
none: 12+12 = 24

Min and max timings for some instructions:
mov A, A -> 4+14 + (1..12)*2 = 20..42
inv A -> 4+10 + (1..12) = 15..26
neg A -> 4+12 + (1..12) = 17..28
abs A -> 4+(12..14) + (1..12) = 17..30

A<0 : mov A, A; jlt : (20..42)+(12..14) = (32..56)
A<=0 : mov A, A; jlt; jeq : (20..42)+(14..26) = (34..68)
A==0 : mov A, A; jeq : (20..42)+(12..14) = (32..56)
A!=0 : mov A, A; jeq : (20..42)+(12..14) = (32..56)
A>0 : mov A, A; jeq : (20..42)+(12..14) = (32..56)
A>=0 : mov A, A; jlt; jeq : (20..42)+(14..26) = (34..68)

Proposed optimizaitons

A<0 : inv A; jgt; jeq : (15..26)+(14..26) = (29..52)
A<=0 : neg A; jgt; jeq : (17..28)+(14..26) = (31..54)
A==0 : neg A; jeq : (17..28)+(12..14) = (29..42)
A!=0 : neg A; jeq : (17..28)+(12..14) = (29..42)
A>0 : neg A; jlt : (17..28)+(12..14) = (29..42)
A>=0 : inv A; jlt : (15..26)+(12..14) = (27..40)

A<0 : abs A; jlt : (17..30)+(12..14) = (29..44)
A<0 : neg A; jgt : (17..28)+(12..14) = (29..42)