Why does `True == False is False` evaluate to False? [duplicate]

I get some rather unexpected behavior on an expression that works with == but not with is:

>>> (True == False) is False
True
>>> True == (False is False)
True
>>> True == False is False
False
>>> id(True)
8978640
>>> id(False)
8978192
>>> id(True == False)
8978192
>>> id(False is False)
8978640

Because in fact that's a chained comparison, so

True == False is False

is equivalent to

(True == False) and (False is False)

This can be surprising in this case, but lets you write 1 <= x < 4 unlike in other languages like C.

From the docs:

x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

In your case True == False is False is equivalent to True == False and False is False as the first condition is False so it short-circuits and return False.

>>> dis.dis(lambda : True == False is False)
  1           0 LOAD_GLOBAL              0 (True)
              3 LOAD_GLOBAL              1 (False)
              6 DUP_TOP             
              7 ROT_THREE           
              8 COMPARE_OP               2 (==)
             11 JUMP_IF_FALSE_OR_POP    21          <---------this step
             14 LOAD_GLOBAL              1 (False)
             17 COMPARE_OP               8 (is)
             20 RETURN_VALUE        
        >>   21 ROT_TWO             
             22 POP_TOP             
             23 RETURN_VALUE  

From the documentation:

5.9. Comparisons

Unlike C, all comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation. Also unlike C, expressions like a < b < c have the interpretation that is conventional in mathematics:

comparison    ::=  or_expr ( comp_operator or_expr )*
comp_operator ::=  "<" | ">" | "==" | ">=" | "<=" | "<>" | "!="
                   | "is" ["not"] | ["not"] "in"