Why does the following evaluates to False in Python?
6==(5 or 6)
False
'b'==('a' or 'b')
False
Why does the following evaluates to False in Python?
6==(5 or 6)
False
'b'==('a' or 'b')
False
 
    
    The first expression evaluates (5 or 6) first, which evaluates to 5 because 5 is truthy. 5 is NOT equal to 6 so it returns False.
The second expression evaluates ('a' or 'b') first, which evaluates to 'a' for the same reason as above. 'a' is NOT equal to 'b' so it returns False.
A good example to explain this would be to try to put a falsey value as the first part of the or expression like 6 == ([ ] or 6) or 6 == (None or 6). Both of these will return true because the or statement will evaluate to the truthy value (in each case it is 6).
There are a couple of ways to create the statement I think you want. The first would be to use in like 6 in (6,5). The second way would be to expand your boolean expression to read like this (6 == 5) or (6 == 6). Both of these will return True.
 
    
    The condition within parentheses is evaluated first. So
6==(5 or 6)
evaluates to
6==(5) # since 5 != 0 (True), the second operand is not evaluated
which is False. Same goes for the second one. ('a' != None)
 
    
    Try:
>>>6 == (6 or 5)
True
What's going on? Try:
>>5 or 6
5
I'm not totally sure why "5 or 6" evaluates to 5, but I would just try writing that expression differently.
 
    
    But there is a caveat:
>> 5 == (0 or 5)
True
Because 0 is not truth-worthy, hence the bracket result in 5.
