Why does Python run the line of code 'if p == 2 or 3' even when the input p is equal to 1? Can someone point out the mistake?

Question

from math import sqrt as s
p = int(input("Enter your number: "))
list = [p]
if p>=4:
for i in range(2,int(s(p))):
if p % i == 0:
print(p,"is not a prime number")
list[:] = []
break
else:
continue
if p in list:
print(p,"is a prime number")
if p == 1:
print('1 is neither prime nor composite')
if p == 2 or 3:
print('{} is a prime number'.format(p))

Answer 1

to expand on Jan Markus explanation: python sees p == 2 or 3 as (p == 2) or 3 and since 3 is not zero, 3 is alway interpreted as true so you have (p==2) or true, which is always true

Answer 2

You could also use:
if p in (2, 3):

Answer 3

Thanks everyone, that was a silly mistake on my part