Line of code not working as expected

A Haskell beginner encountered unexpected behavior in a conditional expression due to operator precedence, where `x && y == False` was parsed as `x && (y == False)` instead of `(x && y) == False`. The fix was to parenthesize the conjunction, and the community also suggested using `not (x && y)` and removing redundant parentheses around string arguments to `print`.

i have a line of code which doesn’t work as i expect it too: parentspermission x y = if x && y == False then print “no” else print “yes” when i run parentspermission False False in ghci it gives me “yes” instead of “no” as it should be when false && false is false this may not be worth mentioning but i am a beginner to haskell ive had some experience in python before and i am against using ai to learn haskell, i want to do it the old school way a little addition is that when i ran parentspermission True False i got “no” as expected but when i ran parentspermission False True i somehow got “yes” Check the precedences of && and == . Neither function is magical in the language. my understanding of this is the order i place && and ==, i would have usually placed a singlar equal sign after x && y but this isnt possible as a single equal sign is used to define functions. The fix was to put x && y in parenthesis like so, x && y == False. Noting that you are new to Haskell, you do not need around the "no" and "yes" values. To apply a function f :: a - b to x :: a , you write f x . Accordingly, you can write print "no" . Also, here, if you want Haskell code to be presented nicely, you can put it between ‘fences’ each on their own lines before and after the code ~~~haskell opening fence and ~~~ closing fence . For example: php parentspermission :: Bool - Bool - IO parentspermission x y = if x && y == False then print “no” else print “yes” A second also: print "no" is equivalent to putStrLn show "no" . As "no" is a String , it may be that you do not actually want to output show "no" but simply "no" . That is because: show "no" == "\"no\"" True where \" is an escaped double quote character in a Haskell string . I’d also add to this that I’d probably usually write if not x && y then ... else ... instead of x && y == False in the conditional. Edit: Also, welcome to Haskell Building on @slow-dive /u/slow-dive 's suggestion, there is a helpful non-AI tool named hlint : which can help you learn by thinking through its suggestions. For example, if you put your code in a module: php module MyModule parentspermission where parentspermission :: Bool - Bool - IO parentspermission x y = if x && y == False then print "no" else print "yes" and command hlint MyModule , it offers up but in colour : MyModule.hs:6:28-44: Suggestion: Redundant == Found: x && y == False Perhaps: not x && y MyModule.hs:6:57-62: Warning: Redundant bracket Found: "no" Perhaps: "no" MyModule.hs:6:75-81: Warning: Redundant bracket Found: "yes" Perhaps: "yes" 3 hints Thx i will check this out