x = True;?

Question

x = True;
y = False;
z = False;
if (not x or y):
print("1")
elif(not x or not y and z):
print("2")
elif(not x or y or not y and x):
print("3")
else:
print("4")
OUTPUT: 3

#Let's evaluate True to 1 and false to 0. So if not 1 or 0 print 1 (cannot be so as could be "or"), If not 1 or not 0 and 0 print 2 (nothing is disqualified, so I guess this is a no). elif not 1 or 0 or not 0 and 1 print three (This comes up as the correct answer but completely confused about why? Is it because one "or" zero vs. not zero "and" one? Any ideas? Thanks to all.

Answer 1

Let's get something clear first.

True = 1, False = 0.
not True = False,
not False = True.

"and" => evaluates to true ONLY IF BOTH operand are True.

"or" => evaluates to True if AT LEAST 1 of the operands is true .

Now, Looking at statement 3
===≠====≠======≠=====
elif((not x or y) or (not y and x))

I've put parentheses to make it clearer.
Now let's evaluate each parentheses one by one.

First:: (not x or y) = (False or False )
which results in False since none of the operand is True.

Second:: (not y and x) = (True and True)
Which results in True since at least one of the operand is True..
Back to the main parentheses

elif((not x or y) or (not y and x)) => (False or True )
Which results in True, therefore 3 is printed

Answer 2

Adding to D'Lite's answer, the "and" operator has higher precedence than the "or" operator.

Answer 3

Thanks DL and Diego👍🏻