python if statement with multiple conditions, and or operator precedence

my original code had no parenthesis around the ... or ... parts of the statements and nothing worked as intended.

The use of -a printed -a (not wanted) and all arguments twice.

The use of -b gave the 2nd if statement of "error: enter filename"

I was referred to this link http://ift.tt/2zuSt6O and thought I understood operator preference.

Thought the brackets would make it check for ==2 and -a, then ==2 and -b and so on.

#!/usr/bin/env python3

import sys
if len(sys.argv) ==1:
    print ("\nerror: please enter at least one filename\n")
    sys.exit()

if len(sys.argv) ==2 and (sys.argv[1]== "-a" or sys.argv[1] == "-b"):
    print ("\nerror: please enter at least one filename after -a or -b\n")
    sys.exit()


if len(sys.argv) >=2 and (sys.argv[1] != "-a" or sys.argv[1] != "-b"):
    for arg in sys.argv[1:]:
        print(arg)


if len(sys.argv) >=2 and (sys.argv[1] == "-a" or sys.argv[1] == "-b"):
    for name in sys.argv[2:]:
        print(name)

However now the use of -a or -b gives the original -a result (prints the -a / -b and also prints all arguments twice).

The use of break stops the repeat but I have no idea why the -a / -b continues to be printed. Is the use of slice [2:] not enough?

Thank you for your help.