if statement gets triggered depending on spelling [duplicate]

I was trying to write a basic calculator (only integers and +, * - operator) in python without importing any modules. Therefore I wrote a function that checks the syntax first and defined all illegal cases.

openingParenthesis = '('
closingParenthesis = ')'

def checkSyntax(self, expression):
    tokens = expression.split()
    counterOpeningParenthesis = 0
    counterClosingParenthesis = 0

    if tokens[0] in ['+', '*', ')'] or tokens[-1] in ['+', '*', '(']:
        return False

    for i in range(0, len(tokens)):
        if tokens[i] == openingParenthesis:
            counterOpeningParenthesis += 1

        elif tokens[i] == closingParenthesis:
            counterClosingParenthesis +=1                

    if counterOpeningParenthesis != counterClosingParenthesis:
        return False

    for i in range(0, len(tokens)-1):

        if tokens[i] == '(' and tokens[i+1] == ')':
            return False

        elif tokens[i] == ('+' or '*') and tokens[i+1] == ('+' or '*'):
            return False

        elif tokens[i] == ')' and tokens[i+1] == '(':
            return False

        elif tokens[i] == ')' and tokens[i+1].isdigit():
            return False

        elif tokens[i] == ('+' or '*') and tokens[i+1] == ')':
            return False

        elif tokens[i].isdigit() and tokens[i+1].isdigit():
            return False

        elif tokens[i] == '(' and tokens[i+1] in ['+', '*']:
            return False

        else:
            return True

When I was testing the syntax of the equation '( * ( ( -2 ) ) )' the following if-statement should have detect an error, but it didn't:

       elif tokens[i] == '(' and tokens[i+1] == ('+' or '*'):
           return False

If I change my if-statement to:

        elif tokens[i] == '(' and tokens[i+1] in ['+', '*']:
            return False

it detects the error. Anybody knows why?