prove [(∀a ∈ M : A(a)) ⇒ B] ≡ ∃a ∈ M : (A(a) ⇒ B)
I tried doing it by changing the predicate on the LHS to:
¬A(a) ∨ B
but it ultimatley ended up being wrong. I also tried using the contraposition rule by negating both sides but it also didnt work.
So how can I prove that the RHS = LHS? thx.
edit: forgot to mention that M is a non-empty set edit 2: I was told by the prof that to solve it using the follwing rules:
- ¬(¬A) ≡ A
- ¬(A ∧ B) ≡ ¬A ∨ ¬B ¬(A ∨ B) ≡ ¬A ∧ ¬B
- A ⇒ B ≡ ¬(A ∧ ¬B) ≡ ¬A ∨ B
- A ⇒ B ≡ ¬B ⇒ ¬A
- A ⇒ B ≡ A ∧ ¬B ⇒ K
- (A ⇒ B) ∧ (B ⇒ C) ⇒ (A ⇒ C) ≡ T
- A ⇔ B ≡ (A ⇒ B) ∧ (B ⇒ A)
- (A ⇒ B) ∧ A ⇒ B ≡ T
- A ∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C) A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C)
- (A ∨ B) ∨ C ≡ A ∨ (B ∨ C) (=: A ∨ B ∨ C) (A ∧ B) ∧ C ≡ A ∧ (B ∧ C) (=: A ∧ B ∧ C)
edit 2: my post was closed because there was a "duplicate" post however, that post is asking about the negation of the universal quantifier (⊢(∀→)↔∃(→)) which does not make it the same problem as my post