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When considering a conditional statement, how does quantification work regarding its individual hypothesis and conclusion? The conditional is a statement, but don't its hypothesis and conclusion also have to be quantified? Or can its hypothesis/conclusion be a predicate instead of a statement?

For example, consider the statement, "If $x$ is a real number, then $x^2 \geq 0.$" Its conclusion "$x^2 \geq 0$" isn't quantified so is a predicate rather than a statement? (Or is there some implicit universal set based on the hypothesis?)

Now consider its contrapositive "If $x^2 < 0$, then $x$ is not a real number." Here, neither its hypothesis nor conclusion is quantified?

ryang
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curiousconfused
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2 Answers2

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There is an implicit "for all" in statements of that form. We'd formalize "if $x$ is a real number, then $x^2\ge0$" as $\forall x(x\in\Bbb R\implies x^2\ge0)$. This way every well-formed statement has no free (i.e. unquantified) variables.

I'd say that the hypothesis and conclusion appear "in a quantified context" in which $x$ is bound, so in this context they represent statements and not predicates.

Notice that the variable names matter: $P(x,y)\implies Q(x,y)$ is very different from $P(x,y)\implies Q(y,x)$. Thinking of the conditional as operating on predicates doesn't make sense unless we associate extra "variable coordination information" with the operation.

Karl
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    I think this confirms that indeed the hypothesis and conclusion, standing each on their own, are predicates and not statements. We get a statement only after the two predicates are combined in the conditional and the quantifier is applied to the conditional. – David K Feb 20 '22 at 17:22
  • I think you could say they're open formulas but not predicates, since variable name coordination matters. I edited my answer to elaborate on this. – Karl Feb 20 '22 at 18:54
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Your statements are both implicitly universally quantified, but their component antecedents and consequents are actually all unquantified: $$\forall x\, \big(x\in\mathbb R\implies x^2\ge0\big)\\\forall x\, \big(x^2<0\implies x\not\in\mathbb R\big).$$

Conventionally, $\text‘P(x){\implies}Q(x)\text’$ is read as $$\forall x\,\Big(P(x){\implies}Q(x)\Big);$$ it never means $$\forall x\,P(x)\implies\forall x\,Q(x),$$ since \begin{align} {}&\forall x\,P(x)\implies\forall x\,Q(x)\\ \equiv{}&\forall x\,\exists y\,\Big(P(y){\implies} Q(x)\Big)\\ \not\equiv{}&\forall x\,\Big(P(x){\implies} Q(x)\Big). \end{align}

ryang
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