# Counterexample: $\Gamma \vdash \phi$ implies $\Gamma\vdash\psi$, but not $\Gamma\vdash \phi\to\psi$

Let $$\Gamma$$ be a collection of formula and $$\phi,\psi$$ be two formulas. Consider two statements:

1. $$\Gamma \vdash \phi\Rightarrow\Gamma\vdash\psi$$
2. $$\Gamma\vdash \phi\to\psi$$

2->1 is fairly easy, one step of modus ponens would suffice. But I cannot prove the inverse direction. I doubt that it is not even true, for there is no new information in 1. to start a derivation, and 2. is not tautology; nor does soundness theorem and completeness theorem seem to work.

What I tried to construct a counterexample: I tried to let $$\Gamma$$ be a familiar set of formulas, such as axioms of group theory or PA. But what is true in group theory is usually also a consequence of these axioms, so I doubt more subtle construction is needed.

Is 1->2 true? If not, what is a counterexample?

(This is asked by a friend of mine who is a graduate student in mathematics and have knowledge of some mathematical logic and set theory.)

• @MauroALLEGRANZA But we do not have that from 1. Feb 21, 2022 at 13:46
• @MauroALLEGRANZA But the OP does not assume they have a derivation of $\Psi$ from $\Gamma$. Feb 21, 2022 at 13:59
• Unless I miss something @MauroALLEGRANZA this is false, examples are easy to construct, just let them both be proportional symbols or something like that, and let Gamma be the set of tautologies. Feb 21, 2022 at 14:02
• It may be worth noting that if $\Gamma$ is a complete theory, then 1. and 2. are equivalent. Assume 1. We have $\Gamma\vdash \phi$ or $\Gamma\vdash \lnot \phi$. In the first case, by 1. $\Gamma\vdash \psi$, so $\Gamma\vdash \phi\to\psi$. In the second case, since $\Gamma\vdash \lnot \phi$, we have $\Gamma\vdash \phi\to \psi$. Feb 22, 2022 at 15:04
• Here is a propositional logic analogy of this question, which includes the solution mentioned by above comments. The answer by Z. A. K., however, provides an excellent first order logic example. Feb 26, 2022 at 15:42

Let $$\Gamma$$ consist of the axioms of group theory, and take the sentence $$\forall x. \forall y. xy=yx$$ as your $$\varphi$$.
Then $$\Gamma \vdash (\forall x. \forall y. xy=yx)$$ fails since not-Abelian groups exist. Since the antecedent is false, the implication $$\Gamma \vdash (\forall x. \forall y. xy=yx) \:\:\Rightarrow\:\: \Gamma \vdash (\forall a.\forall b. a=b)$$ is vacuously true.
However, $$\Gamma \vdash (\forall x. \forall y. xy=yx) \rightarrow (\forall a. \forall b. a=b)$$ does not hold, since Abelian groups with more than one element do exist.