Trivial and Vacuous Proofs

We usually consider

\[ P(x) \implies Q(x) \]

• where the variable(s) \(x \in \mathbb{S}\), universal set

Observations:

• \(P(x) \implies Q(x)\) is True if it is true for all \(x \in \mathbb{S}\)
• \(P(x) \implies Q(x)\) is False if it is false for at least one \(x \in \mathbb{S}\).

Next, recall:

1. Observe: If \(Q(x)\) (the conclusion) is always True, then \(P(x) \implies Q(x)\) (the implication) is True.
2. Called: A trivial proof (the hypothesis is irrelevant)

Example:

1. Let \(x \in \mathbb{R}\).
2. If \(x < 0\), then \(x^2+1 > 0\).
   • Observe: \(\forall x \in \mathbb{R}\), we have \(x^2 \geq 0\).
   • So \(x^2+1 \geq 0 +1 > 0\)
   • Meaning the conclusion is always true.
   • So, this result is trivially true.

Example:

1. Observe: If \(P(x)\) (the hypothesis) is always false, then \(P(x) \implies Q(x)\) is true.
2. Called: A vacuous proof (the conclusion is irrelevant)

Example:

1. \(x \in \mathbb{R}\). If \(x^2 - 2x + 2 \leq 0\) then \(x^3 \geq 8\).
2. Proof:
   • Observe \(\forall x \in \mathbb{R}\), we have
     • \(x^2 - 2x + 2 = (x^2 - 2x + (-1)^2) + 2\)
     • \(= (x-1)^2 + 1 \geq 0 + 1\)
     • \(> 0\)
   • Meaning the hypothesis is always false.
   • So this result is vacuously true.

Recall the classic fact: For all sets \(A, \emptyset \subseteq A\). We can now prove this!

• If \(x \in \emptyset\), then \(x \in A\).
• Note that \(x \in \emptyset\) is always false.
• Therefore, this statement is vacuously true!