Prove that [(A - C) × B] ∪ [A × (B - D)] ⊆ (A × B) - (C × D)

Proof:

  Suppose p ∈ [(A - C) × B] ∪ [A × (B - D)]
  Then either (i) a ∈ A - C and b ∈ B, or (ii) a ∈ A and b ∈ B - D exist such that p = (a,b)

  (i) From a ∈ A - C and b ∈ B, we get a ∈ A ∧ b ∈ B ∧ ¬(a ∈ C)
      A fortiori, a ∈ A ∧ b ∈ B ∧ ¬(a ∈ C ∧ b ∈ D)
      Thus p ∈ (A × B) - (C × D)

  (ii) From a ∈ A and b ∈ B - D, we get a ∈ A ∧ b ∈ B ∧ ¬(b ∈ D)
      A fortiori, a ∈ A ∧ b ∈ B ∧ ¬(a ∈ C ∧ b ∈ D)
      Thus p ∈ (A × B) - (C × D)

  Thus [(A - C) × B] ∪ [A × (B - D)] ⊆ (A × B) - (C × D)

# posted by rot13(Unafba Pune) @ 11:50 AM