Can Find(<image> or <image>) be used, what would the syntax be?

You have to make it on your own, since there is no such feature.

-- The easiest way: using sequential exists()
if exists(img1):
   #do something
elif exists(img2):
   # do something else
else:
   print "neither img1 nor img2"

This depends on the sequence and img2 processing will not be done, if img1 is there. If img1 does not exist it will take 3 seconds until img2 will be searched.

--- most sophisticated
# each find action is only done once
# all images are found
# information about what images were found is available
# a max time for whole action is spacified
imgs = [img1, img2, img3]
t = 3 # waiting time
res = [False for img in imgs]

start = time.time()
while time.time()-start < t:
    for i in range(len(imgs)):
        if exists(imgs[i], 0):
            res[i] = True
    if True in res: break
print res

res contains the information, which of the images where found

so if you only want to know, that at least one of the images was there:

if True in res:
    # do something
else:
    # none of the images where found

specific actions depending on a specific image:

if res[0]:
    # special action for image 1

*** comments
- when trying to understand scripts, it is a good idea to look into the docs in parallel
http://sikuli.org/docx/
- you need some basic knowledge of Python language, if you want to do more complex things. Especially the indentation magic is always a hurdle for beginners.

Hi RaiMan or fellow moderators or creators,

Not sure if you are still helping out but I have an important question that has been bugging me for weeks.

How do I compute the following:

WAIT 120 secs for EITHER image A or image B to appear

If A appear, do step 1 (which includes a number of steps)
If B appear, do step 2

There will not be a case where both or neither of them appear, I don't want to set a sleep timer as I want the program to be robust and snappy

@Josh
no need to wait weeks until asking such questions ;-)
TIP: you would get more possible responses, if you post your own question, and not dock to an existing one.

BTW: the concepts for your solution is already sketched in comment #1 and still valid:

the steps, to get the most effective solution:
- evaluate the smallest possible region, where you expect the 2 images to appear (one or two regions, does not matter, minimum size is the goal)
- make a wait loop
- use exists(img, 0) to check appearence

reg1 = #evaluate region for image1
reg2 = #evaluate region for image2
waitUntil = time.time() + 120
while time.time() < waitUntil:
    if reg1.exists(image1, 0):
        # do what is needed
        break
    if reg2.exists(image2, 0):
        # do what is needed
        break