I've been an ArcGIS Python programmer for years and I've never really needed the id() function or the is operator to get real work done. In practice, I only really care about equality (==). Since you are just experimenting, I guess that's reason enough, but it's probably not a rabbit hole you need not delve in.
So here's what happening in the code above.
s = 'a string'
# 1) The print statement is called with a comma, which means each parameter
# will be evaluated, one at a time, from left to right.
# 2) s[3:] produces a new string with an id
# 3) That string is passed to the id function, which returns its memory address
# 4) That memory address goes into a new int object
# 5) The string is no longer needed, so it is freed
# 6) s[5:] produces a new string, reusing free memory
# 7) That string is passed to the id function, which returns its memory address
# 😎 That memory address goes into a new int object
# 9) The string is no longer needed, so it is freed
# Both objects had the same id at the time they were alive, so the
# same id is printed.
print(id(s[3:]), id(s[5:]))
# Similar to above, each parameter is evaluated left to right.
# This time, print only has one parameter, but that parameter
# is the result of an operator. The operator resolves its operands
# from left to right.
# Follow steps 1 through 9, paying attention to 4) and 8).
# Int objects are equal to each other if they have the same value
# so True is returned to the print statement.
print(id(s[3:])==id(s[5:]))
# 1) The print statement is called with one parameter, as evidenced by
# its lack of commas
# 2) The print statement's only parameter is the result of the is operator
# 3) The is operator evaluates its operands from left to right
# 4) s[3:] produces a new string with an id
# Note that this new string is still needed for the is operator, so it
# is not freed. It hangs out in memory
# 6) s[5:] produces a new string with an id
# This new string is also needed for the is operator, so it is not freed either
# 7) The is operator now compares the identity of both living strings,
# sees that they aren't equal, and returns False in both instances.
print(s[3:] is s[5:])
print(s[3:] is s[3:])
Now, to address @DanPatterson 's post:
s = 'a string'
# Since there are no print statements, I'm not sure how he got results
# that are in the comments. Perhaps, if run in a terminal, or at different
# times, the new strings did not reuse recently freed memory.
id(s) # 2625510363760
id(s[:3]) # 2625512614384
id(s[:3]) # 2625512614000
id(s[:3]) # 2625512908528
# 1) The is operator evaluates its operands from left to right
# 2) s[:3] creates a new string with an id
# 3) The id function is called, returning a new int object
# 4) The new string is freed, releasing its memory
# 5) s[:3] creates a new string with an id. This may or may
# not have been created in the same memory space as 1) was
# 6) The id function is called, returning a new int object
# 7) The is operator attempts to establish identity on the
# two ints that were returned, which have not been freed.
# The ints are different objects, so is returns False.
id(s[:3]) is id(s[:3]) # False
# This is actually the easiest to explain. Dan could have
# typed id(10>7), id("h" in "PYTHON".lower()), or id(5!=7)
# and still gotten the same memory address.
# What he is doing is grabbing the id of the result of the
# _equality_ operator. Those results are bools. The values
# True and False are global objects of type bool. Because
# they are objects, they each have an id. Comparator operators
# don't create new instances of type bool, but simply
# use the existing ones.
id(id(s[:3]) == id(s[:3])) # 140711459289192
id(s[:3] == s[:3]) # 140711459289192
