Let \(P(x)\) and \(Q(x)\) be open sentences containing the variable \(x\), and consider the following statements. \[A: \forall x, (P(x) \Rightarrow Q(x)) \quad \text{and} \quad B: (\forall x, P(x)) \Rightarrow (\forall x, Q(x))\]
- Prove that \(A \Rightarrow B\).
- Give an example of open sentences \(P(x)\) and \(Q(x)\) to show that \(B \Rightarrow A\) need not be true.
<1>2. Then \(\forall x\,Q(x)\).
<1>3. Hence \(A\Rightarrow B\).
<1>4. The converse can fail. Take the universe to be \(\mathbb Z\), and define \[ P(x): x=0, \qquad Q(x): x=1. \]
<1>5. For these predicates, \(B\) is true but \(A\) is false.
Solution
1. Assume \(A\), and suppose \(\forall x\,P(x)\).
Let \(x\) be arbitrary. From \(\forall x\,P(x)\) we have \(P(x)\). From \(A\) we have \(P(x)\Rightarrow Q(x)\). Therefore \(Q(x)\). Since \(x\) was arbitrary, \(\forall x\,Q(x)\).
Under the assumption \(A\), <1>1–<1>2 prove the implication \[ (\forall x\,P(x))\Rightarrow(\forall x\,Q(x)), \] which is exactly \(B\).
The antecedent \(\forall x\,P(x)\) of \(B\) is false, so \(B\) is true. But at \(x=0\), the statement \(P(0)\) is true and \(Q(0)\) is false; therefore \(P(0)\Rightarrow Q(0)\) is false, so \(A\) is false.