Proof that $A^* - \{x\} = A^+$ such that $A=\{a,b\}$ and $x \in \{a,b\}^*$ Exercise 1.1 (2) from Problem Solving in Automata, Languages, and Complexity by Du and Ko. P.7. the question is as following:


  Let A be a langauge over {a,b} and $x \in {a,b}^*$. Find necessary and sufficient conditions in terms of x and A for the equation:
  $$A^* - \{x\} = A^+$$


thus, by substation:
$$\{a,b\}^* - \{\{a,b\}^*\} = \{a,b\}^+$$

Now this the LHS should equal to $\{a,b\}^*$ because if we have two sets A and B, then $A \cap B = \phi$, then $A-B=A$. Now where is wrong?!!

I'm also trying to prove from RHS since this is equality to proof should be from both direction.
Your problem is that in going from $A^*-\{x\}$ to $\{a,b\}^*-\{\{a, b\}^*\}$ you went a step too far. In the first expression, $x$ is a single string over the alphabet. Replacing that with the general expression $\{\{a,b\}^*\}$ is incorrect, if for no other reason that $\{\{a,b\}^*\}$ is a set consisting of a set of words and not a single word.

To complete this answer, see newbie's comment on your original post.