The Direct Strategy for Proving Implications Involving Quantifiers

Question

I want to prove a statement of the form ∀x ∈ X [ P(x) ⇒ Q(x) ], where P(x) and Q(x) are statements about x in the set X.

The purpose of this post is to make sense of a strategy used to prove the above in simple English.

This means that words such as "arbitrary," and "implies" will be conveyed using commonly used English words. The following is my attempt:

∀x ∈ X [ P(x) ⇒ Q(x) ] is defined as for every x in X, if P(x) is true, then Q(x) is true.

Definition of p implies q(if p, then q): In every case where p is true, q must also be true.

By this definition, in every case where P(x) is true for x in X, Q(x) must also be true. In other words, the truth of P(x), where x is in X, guarantees the truth of Q(x).

This is a strong claim because it says that the truth of Q(x) logically follows every time P(x) is true for x in X.

To prove this, we let x be any element in X, and assume that P(x) is true. Then, we logically arrive at Q(x).

Once this is done, no matter which element in X we take x to be, whenever P(x) is true, Q(x) is true. This proves that for every x in X, if P(x) is true, then Q(x) is true.

---

The definition of p implies q has the essence of causal implication, where p and q are related. This was intended because I want to focus on proofs in mathematics, where the relationships are causal.

This is supported by what Keith Devlin says on pages 32 and 33 of Introduction to Mathematical Thinking:

The conditional and bi-conditional only differ from implication and equivalence in situations that do not arise in the course of normal mathematical practice.

Is my thinking precise when it comes to the strategy for proving implications?

Answer 1

The definition of p implies q has the essence of causal implication, where p and q are related.

Do not conflate implication with causality. In mathematics based on classical logic, there is no notion of causality.

Suppose, for example, that $Q$ can be formally derived from $P$. This rules out only the possibility that both $P$ is true and $Q$ is false. We can write:

$\neg (P \land \neg Q)~~~~~$ or equivalently $~~~~~P\implies Q$

Example: Consider the implication, it is raining ($R$) implies it is cloudy ($C$).

$R \implies C$

This does not mean that rain causes cloudiness. Or that it is always cloudy when it is raining (e.g. so-called sunshowers). In classical propositional logic, it means only that it is not the case that it is both raining and not cloudy.

$\neg (R \land \neg C)$

Is my thinking precise when it comes to the strategy for proving implications?

To formally prove $P\implies Q$, you can do any of the following:

1. Derive $Q$ from $P~~~$ (Direct proof)
2. Derive $\neg P$ from $\neg Q~~~$ (Proof by contrapositive)
3. Derive a contradiction from $P$ and $\neg Q~~~$ (Proof by contradiction)
4. Prove $\neg P ~~~ $ (Vacuous truth, or principle of explosion, or arbitrary consequent)
5. Prove $Q ~~~ $ (Arbitrary antecedent)

---

EDIT

While proving ∀x ∈ X [ P(x) ⇒ Q(x) ], why do we start with x being any element in X, and assume that P(x) is true? Is there a reason to it? (From comment)

You really have 2 nested implications:

$\forall a: [a \in X \implies [P(a)\implies Q(a)]]$

To prove it, you could start by introducing the premise (a what-if statement) $x\in X$. Then you can introduce a 2nd premise: $P(x)$.

Then, by direct proof, you might prove $Q(x)$. Then $P(x)\implies Q(x).$

Finally, you can generalize using a universal quantifier to obtain the required result.

(1) $x\in X ~~~~~$ (Premise)

(2) $P(x) ~~~~~$ (Premise)

$~~~\vdots$

(n) $~Q(x)$

(n+1) $~P(x) \implies Q(x) ~~~~~ $(Conclusion 2)

(n+2) $~x\in X \implies [P(x) \implies Q(x)] ~~~~~ $ (Conclusion 1)

(n+3) $~\forall a: [a \in X \implies [P(a)\implies Q(a)]]~~~~~ $ (Generalize)