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How chain of `and` operators works in python?

In my program encountered with this:
>>> True and True and (3 or True)
3
>>> True and True and ('asd' or True)
'asd'
while I expected to get some boolean value depending on the result in brackets. So if I try comparisons like these (0 or True) or ('' or True) python will return True, which is clear because 0 and '' equivalent to False in comparisons.
Why doesn't python return boolean value by converting 3 and 'asd' into True?

From https://docs.python.org/3/library/stdtypes.html:
Important exception: the Boolean operations or and and always return
one of their operands
The behavior can be most easily seen with:
>>> 3 and True
True
>>> True and 3
3
If you need to eliminate this behavior, wrap it in a bool:
>>> bool(True and 3)
True
See this question
As Reut Sharabani, answered, this behavior allows useful things like:
>>> my_list = []
>>> print (my_list or "no values")

Comparison operators and 'is' - operator precedence in python?

So I was looking at some code online and I came across a line (at line 286):
if depth > 0 and best <= -MATE_VALUE is None and nullscore > -MATE_VALUE:
The part I had trouble understanding was the best <= -MATE_VALUE is None.
So I fired up the interpreter to see how a statement such as value1 > value2 is value3 work. So I tried
>>> 5 > 2 is True
False
>>> (5 > 2) is True
True
>>> 5 > (2 is True)
True
My Question
Why is 5 > 2 is True not True? And how do these things generally work?
Thanks.

You're seeing python's operator chaining working
5 > 2 is True
Is equivalent to
5>2 and 2 is True
You can see this in that
>>> 5>2 is 2
Returns True.

First, 5 > 2 is True is equivalent to (5 > 2) and (2 is True) because of operator chaining in python (section 5.9 here).
It's clear that 5 > 2 evaluates to True. However, 2 is True will evaluate to False because it is not implicitly converted to bool. If you force the conversion, you will find that bool(2) is True yields True. Other statements such as the if-statement will do this conversion for you, so if 2: will work.
Second, there is an important difference between the is operator and the == operator (taken from here):
Use is when you want to check against an object's identity (e.g.
checking to see if var is None). Use == when you want to check
equality (e.g. Is var equal to 3?).
>> [1,2] is [1,2]
False
>> [1,2] == [1,2]
True
While this does not have an immediate impact on this example, you should keep it in mind for the future.

Why do 'and' & 'or' return operands in Python?

I'm going through the LPTHW and I came across something I cannot understand. When will it ever be the case that you want your boolean and or or to return something other than the boolean? The LPTHW text states that all languages like python have this behavior. Would he mean interpreted vs. compiled languages or duck typed vs static typed languages?
I ran the following code:
>>> False and 1
False
>>> True and 1
1
>>> 1 and False
False
>>> 1 and True
True
>>> True and 121
121
>>> False or 1
1
>>> False or 112
112
>>> False or "Khadijah"
'Khadijah'
>>> True and 'Khadijah'
'Khadijah'
>>> False or 'b'
'b'
>>> b = (1, 2, "K")
>>> b
(1, 2, 'K')
>>> False or b
(1, 2, 'K')
>>>
Please help me understand whats going on here.
According to the documentation: http://docs.python.org/2/library/stdtypes.html
Operations and built-in functions that have a Boolean result always return 0 or False for false and 1 or True for true, unless otherwise stated. (Important exception: the Boolean operations or and and always return one of their operands.)
According to LPTHW: http://learnpythonthehardway.org/book/ex28.html
Why does "test" and "test" return "test" or 1 and 1 return 1 instead of True?
Python and many languages like it return one of the operands to their boolean expressions rather than just True or False. This means that if you did False and 1 you get the first operand (False) but if you do True and 1 your get the second (1). Play with this a bit.

I think you're somehow confused about what the docs says. Take a look at these two docs sections: Truth Value Testing and Boolean Operators. To quote the last paragraph on the fist section:
Operations and built-in functions that have a Boolean result always return 0 or False for false and 1 or True for true, unless otherwise stated. (Important exception: the Boolean operations or and and always return one of their operands.)
As you can see, you're right about operations and built-in functions but see the Important exception part, it is well stated that the Boolean operators will return one of their operands.
Now, what they can return depends hardly on the operator's short circuit logic. For or operator, it will return the first truthy value in the expression, since when it finds one, the whole expression is true. In case of every operand being falsey, or will return the last operand, meaning that it iterated over every one of them not being able to find a truthy one.
For and operator, if the expression is true, it will return the last operand, if the expression is false, it will return the first falsey operand. You can read more about Short Circuit Evaluation at the Wikipedia Page.
You have a lot of examples in your question, let's analyze some of them:
>>> False and 1 # return false (short circuited at first falsey value)
False
>>> True and 1 # return 1 (never short circuited and return the last truthy value)
1
>>> 1 and False # return false (short circuited at first falsey value, in this case the last operand)
False
>>> 1 and True # return True (never short circuited and return the last truthy value)
True
>>> True and 121 # return 121 (never short circuited and return the last truthy value)
121
>>> False or 1 # return 1 (since the first operand was falsey, or kept looking for a first truthy value which happened to be the last operator)
1
>>> False or 112 # return 112 for same reason as above
112
>>> False or "Khadijah" # return "Khadijah" for same reason as above
'Khadijah'
>>> True and 'Khadijah' # return "Khadijah" because same reason as second example
'Khadijah'
I think this should make a point. To help you further understand why this is useful, consider the following example:
You have a function that randomly generate names
import random
def generate_name():
return random.choice(['John', 'Carl', 'Tiffany'])
and you have a variable that you don't know if it has assigned a name yet so instead of doing:
if var is None:
var = generate_name()
You can do oneliner:
var = var or generate_name()
Since None is a falsey value, or will continue its search and evaluate second operand, this is, call the function ultimately returning the generated name. This is a very silly example, I have seen better usages (although not in Python) of this kind of style. I couldn't come out with a better example right now. You can also take a look at this questions, there are very useful answers on the topic: Does Python support short-circuiting?
Last but not least, this has nothing to do with static typed, duck typed, dynamic, interpreted, compiled, whatever language. It's just a language feature, one that might come handy and that is very common since almost every programming language I can think of provide this feature.
Hope this helps!

One would want and and or to evaluate to an operand (as opposed to always evaluating to True or False) in order to support idioms like the following:
def foo(self):
# currentfoo might be None, in which case return the default
return self.currentfoo or self.defaultfoo()
def foobar(self):
# foo() might return None, in which case return None
foo = self.foo()
return foo and foo.bar()
You can of course question the value of such idioms, especially if you aren't used to them. It's always possible to write equivalent code with an explicit if.
As a point against them, they leave some doubt whether the full range of falsey values is possible and intentionally accounted for, or just the one mentioned in the comment (with other falsey values not permitted). But then, this is true in general of code that uses the true-ness of a value that might be something other than True or False. It occasionally but rarely leads to misunderstandings.

This has to do with the way short circuit effect is implemented in Python.
With and (remember True and X = X), the result of the right expression is pushed into the stack, if it's false, it gets poped immediately, else the second expression is poped:
>>> import dis
>>>
>>> dis.dis(lambda : True and 0)
1 0 LOAD_CONST 2 (True)
3 JUMP_IF_FALSE_OR_POP 9
6 LOAD_CONST 1 (0)
>> 9 RETURN_VALUE
>>>
>>> True and 0
0
>>> 0 and True
0
>>>
similar to:
def exec_and(obj1, obj2):
if bool(obj1) != False:
return obj2
return obj1
With or, if the first expression is true, it gets popped immediately. If not, the second expression is poped, now the result really depends on the second.
>>> dis.dis(lambda : 0 or False)
1 0 LOAD_CONST 1 (0)
3 JUMP_IF_TRUE_OR_POP 9
6 LOAD_CONST 2 (False)
>> 9 RETURN_VALUE
>>>
>>> True or 0
True
>>> 1 or False
1
>>> False or 1
1
>>>
similar to:
def exec_or(obj1, obj2):
if bool(obj1) != True:
return obj2
return obj1

Consider the following use case:
element = dict.has_key('foo') and dict['foo']
Will set element to dict['foo'] if it exists, otherwise False. This is useful when writing a function to return a value or False on failure.
A further use case with or
print element or 'Not found!'
Putting these two lines together would print out dict['foo'] if it exists, otherwise it will print 'Not found!' (I use str() otherwise the or fails when element is 0 (or False) because that s considered falsey and since we are only printing it doesn't matter)
This can be simplified to
print dict.has_key('foo') and str(dict['foo']) or 'Not found!'
And is functionally equivalent to:
if dict.has_key('foo'):
print dict['foo']
else:
print 'Not found!'

Python Bool and int comparison and indexing on list with boolean values

Indexing on list with boolean values works fine.
Though the index should be an integer.
Following is what I tried in console:
>>> l = [1,2,3,4,5,6]
>>>
>>> l[False]
1
>>> l[True]
2
>>> l[False + True]
2
>>> l[False + 2*True]
3
>>>
>>> l['0']
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: list indices must be integers, not str
>>> type(True)
<type 'bool'>
When I tried l['0'] it printed error that int type expected in indices and that is obvious.
Then, even the type of 'True' and 'False' being Bool, indexing on the list works fine and automatically converts it to int type and performs the operation.
Please explain what is going on internally.
I am posting question for the first time, so please forgive me for any mistake.

What's going on is that booleans actually are integers. True is 1 and False is 0. Bool is a subtype of int.
>>> isinstance(True, int)
True
>>> issubclass(bool, int)
True
So it's not converting them to integers, it's just using them as integers.
(Bools are ints for historical reasons. Before a bool type existed in Python, people used the integer 0 to mean false and 1 to mean true. So when they added a bool type, they made the boolean values integers in order to maintain backward compatibility with old code that used these integer values. See for instance http://www.peterbe.com/plog/bool-is-int .)
>>> help(True)
Help on bool object:
class bool(int)
| bool(x) -> bool
|
| Returns True when the argument x is true, False otherwise.
| The builtins True and False are the only two instances of the class bool.
| The class bool is a subclass of the class int, and cannot be subclassed.

Python used to lack booleans, we just used integers, 0 for False and any other integer for True. So when booleans were added to the language, the values False and True, can be treated as the integer values 0 and 1 still by the interpreter, to help backwards compatibility. Internally, bool is a sub-class of int.
In other words, the following equations are True:
>>> False == 0
True
>>> True == 1
True
>>> isinstance(True, int)
True
>>> issubclass(bool, int)
True
and as you found out:
>>> True * 3
3
This doesn't extend to strings however.

...Booleans are a subtype of plain integers.
Source.
As you can see, False is 0 and True is 1.

The Python source documentation does not mention directly that all non-zero integers are evaluate to True when passed to an if statement, while only zero evaluates to False. You can prove it to yourself with the following code in Python:
for test_integer in range(-2, 3, ):
if not test_integer:
print('{} evaluates to False in Python.'.format(test_integer))
else:
print('{} evaluates to True in Python.'.format(test_integer))
>>>-2 evaluates to True in Python.
-1 evaluates to True in Python.
0 evaluates to False in Python.
1 evaluates to True in Python.
2 evaluates to True in Python.
Try it for as far on either side of zero as you want; this code only shows for -2, -1, 0, 1, and 2 inclusive.

Python max() and min() values for bool

In python interpreter,
min(True,False)==False
max(True,False)==True
is assured by design?

True is equal to 1 and False is 0

It seems, at least in CPython, bool subclasses int. Therefore, you can do:
>>> abs(False)
0
>>> abs(True)
1
and:
>>> False < True
True
>>> True > False
True
I guess max and min work on the comparison operator:
>>> cmp(False, True)
-1
>>> cmp(True, False)
1
>>> cmp(False, False)
0
>>> cmp(True, True)
0

In python 2.x, this is not guaranteed, as you can overwrite True and False:
>>> False = 23
>>> max(True, False)
23
But if you do not assign to True or False, it is guaranteed by Language Design that Booleans subclass int with values 0, 1, yes. (and in py3, True and False are reserved words, so you cannot do the above)

According to the doc,
In numeric contexts (for example when used as the argument to an arithmetic operator), they behave like the integers 0 and 1, respectively.
So, yes, it is assured.