Solving a Mystery, Case by Case

Some of the proofs we have seen before were pretty direct. When you have P(x) ⇒ Q(x), there is little that can go wrong~! But things do become a little bit more complicated when we encounter ∧. This conjunction symbol indicates that, if we are to prove something, we need to show that both sides are true. When you find a conjunction on the antecedent, such as having P(x) ⇒ Q(x) were P(x) ⇔ R(x) ∧ S(x), you can just leave it as the assumption.

For example:

∀ x ∈ ℕ, R(x) ∧ S(x) ⇒ Q(x)

Would lead us to:
Assume x ∈ ℕ
    Assume R(x) ∧ S(x)
         ....
    Then R(x) ∧ S(x) ⇒ Q(x)
Then ∀ x ∈ ℕ, R(x) ∧ S(x) ⇒ Q(x)

This is fairly easy, and makes our lives' easier! Nonetheless, there will be plenty of times when we will want to disprove a conjunction. To do this, we will have to work on different cases. Let's look at the converse of the previous statement, and let's assume it is true:

∀ x ∈ ℕ, Q(x) ⇒ R(x) ∧ S(x)

This would lead us to:
Assume x ∈ ℕ
    Assume Q(x)
         Then...
             Assume... #case 1 that leads to proving R(x)
                 Then...
             Assume... #case 2 that leads to proving S(x)
                 Then...
         Then...
    Then Q(x) ⇒ R(x) ∧ S(x)
Then ∀ x ∈ ℕ, Q(x) ⇒ R(x) ∧ S(x)

As you can see, separate cases spawn while we are working on this proof. The important thing is to remember that these cases need to lead to proving R(x) and S(x) and need to be within context of Q(x). That sounds like a lot of work, but many times it is easy to see the light!