As the title implies, I'm totally lost with proofs. I'm currently writing the section that is called the "Scratch Work" section, where we have to write a sequence of steps, outlining our logic, before we write our actual proof. So I'm middle of working on this and it is confusing the heck out of me. This is what I have, but I am not confident in any of the steps at all.
Original question:
Let A, B and C be sets such that A and B \ C are disjoint. Prove that A ∩ B ⊆ C.
Givens: Goals:
1) Sets A and B\C are disjoint A ∩ B ⊆ C
Let x be arbitrary
2) ∀x[(x ∈ A) ∧ (x ∉ (B\C))] ∀x(x ∈ A ∩ B → x ∈ C)
Universal Instantiation
3) x ∈ A ∧ x ∉ B\C x ∈ A ∩ B → x ∈ C
Applying deduction method
4) x ∈ A ∩ B(new premise after deduction) x ∈ C
5) x ∈ A ∧ x ∈ B (set intersection def, 4)
6) x ∈ A (sim, 3)
7) x ∉ B\C (sim, 3)
8) x ∈ B (sim, 5)
I have no idea if the switching from the first given and goal to the second given and goal is correct (switching to quantifier statements) and I dont know how to get to the goal. Any nudge in the right direction would be greatly appreciated.
EDIT - Solution proposal:
Scratch Work:
Givens: Goals:
1) Sets A and B\C are disjoint A ∩ B ⊆ C
A ∩ (B\C) = ∅
Let x be arbitrary
2) ∀x(x ∉ A ∩ (B\C)) ∀x(x ∈ A ∩ B → x ∈ C)
Universal Instantiation
3) x ∉ A ∩ (B\C) x ∈ A ∩ B → x ∈ C
Applying deduction method
4) x ∈ A ∩ B (new premise after deduction) x ∈ C
5) x ∉ A ∧ (B\C) (set intersection def, 3)
6) x ∉ (B\C) (sim, 5)
7) x ∈ A ∧ B (set intersection def, 4)
8) x ∈ B (sim, 7)
9) x ∈ C (definition of set difference, 6 and 8)
Let $x\in A\cap B$, then we have $x\in A$ and $x\in B$. As we have $A$ and $B-C$ are disjoint it implies that $A\cap(B-C)=\emptyset$. That means as $x$ is in both $A$ and $B$ that $x$ must be in $C$ as $B-C=\{x:x\in B, x\notin C\}$ to as otherwise it cannot be equal to the empty set. Ergo $A\cap B\subseteq C$.