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I have a little question on P. Cohen's book "Set Theory and the Continuum Hypothesis". On page 10, the following inference rule (F) of a FOL deduction system is given:

Let $A(x)$ be a formula with one free variable $x$. Suppose $B$ is a formula not involving a constant symbol $c$ and $x$. Then $A(c)\rightarrow B$ is valid implies $\exists xA(x)\rightarrow B$ is valid.

I think there is a mistake since we may construct a model where $\exists xA(x)$ is true while $A(c)$ and $B$ are false. Thus we have $A(c)\rightarrow B$ is true but $\exists xA(x)\rightarrow B$ is false. Could I miss something here?

Bertrand Haskell
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    You cannot "validate" it with a constant whatever: consider the FO language of arithmetic with constant $0$. We have that $(0 > 0) \to (0 \ne 0)$ is True while $\exists x (x > 0) \to (0 \ne 0)$ is False. – Mauro ALLEGRANZA Feb 01 '22 at 10:37
  • @MauroALLEGRANZA thanks for the response, I'll make a note in my book. – Bertrand Haskell Feb 01 '22 at 10:44
  • The counter-example above show that the Rule is not "sound" in the usual way, i.e. preserving truth. The trick is in the previous paragraph the Rule; obviously $(0 > 0) \to (0 \ne 0)$ is not valid. If we restrict the Rule to "preserve validity" then a formula $A(c)$ cannot be False, because the only way to falsify it is to have an empty domain. – Mauro ALLEGRANZA Feb 01 '22 at 11:05
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    See also this post – Mauro ALLEGRANZA Feb 01 '22 at 11:46
  • @MauroALLEGRANZA In the way you gave (when we use $A\rightarrow B$ instead of $A(c)\rightarrow B$ given $B$ contains no free $x$), the rule is totally clear. – Bertrand Haskell Feb 01 '22 at 11:56

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