1

The soundness theorem for first-order logic can be stated as follows: If $\Delta\vdash\varphi$, then $\varphi$ is valid in all structures (for the given language) in which all formulas in the set of formulas $\Delta$ are valid ("valid" meaning: satisfied by all evaluations).

In a course I attend we were writing this as follows ("D" being a structure, meaning a domain and a realization of the symbols of the given language):

$$\Delta\vdash\varphi\Rightarrow\forall{D}(\forall\psi\in\Delta({D}\vDash\psi)\Rightarrow{D}\vDash\varphi)$$

Now the consequence relation was defined as follows ("e" being an evaluation for a given formula and structure):

$$\Delta\vDash\varphi\Rightarrow\forall{D}\forall{e}(\forall\psi\in\Delta({D}\vDash\psi[e])\Rightarrow{D}\vDash\varphi[e])$$

Then it was stated, that the consequence (the part right of the $\Rightarrow$) of the soundness theorem is not generally equivalent with the definition of the consequence relation, it is so only when all formulas we are concerned with are sentences (i.e. containing no free variables). We were rewriting those as follows:

$$(st):\forall{D}(\color{red}{\forall{e}}\forall\psi\in\Delta({D}\vDash\psi[e])\Rightarrow\color{red}{\forall{e}}{D}\vDash\varphi[e])$$ $$(cn):\forall{D}\color{red}{\forall{e}}(\forall\psi\in\Delta({D}\vDash\psi[e])\Rightarrow{D}\vDash\varphi[e])$$

My question now is: How do (st) and (cn) actually differ? One example we were given to show the difference is the following:

$$(1):\{P(x)\}\nvDash\forall{yP(y)}$$

It was said that in this case (st) is met, but (cn) is not. I don't quite understand it: What does it even mean to say, that some formula $\varphi$ is a consequence of some open formula $\psi$? According to the above definition it means, that there can be no evaluation for which $\psi$ is valid and $\varphi$ not. So in our case (1) I see why it is not a consequence: I can choose a structure for which there is one element a in P and one element b that is not in P, so for the evaluation a $\rightarrow$ x our definition of a consequence is not met (since $\forall{yP(y)}$ is not satisfied by this evaluation but $P(x)$ is).

But it was said that (st) is met. How come? I don't quite understand how the change of the all-quantifiers over e (indicated in red above) works here - why is not the same reasoning true about (st) and (1)?

This is the main part of my question and I'm very glad about every answer to it! However I have a bit more context, in which I am asking myself this question and if there's anyone able to help with that I'd appreciate it very much. We are supposed to show the following for any set of formulas $\Delta$ and formulas $\psi$ and $\varphi$:

$$(2):\Delta,\psi\vDash\varphi\iff\Delta\vDash\psi\rightarrow\varphi$$

If we are concerned with theories (no open formulas), I guess both directions of the equivalence can be shown using the completeness theorem, the deduction theorem and the soundness theorem. For the right direction:

$$\Delta,\psi\vDash\varphi\Rightarrow_{Completeness}\Delta,\psi\vdash\varphi\Rightarrow_{Deduction}\Delta\vdash\psi\rightarrow\varphi\Rightarrow_{Soundness}\Delta\vDash\psi\rightarrow\varphi$$

But I have the feeling, that one cannot do the same if $\psi$ and $\varphi$ might be open formulas. How can I prove (2) solely in terms of evaluations and structures? If I understand it correctly, I have to show (for the right direction) that if there can be no evaluation for which $\Delta,\psi$ is valid and $\varphi$ not, then there can be no evaluation for which $\Delta$ is valid and $\psi\rightarrow\varphi$ is not. But I get lost here with the subtleties of the quantification over the evaluation and stuff..

One idea with which I'm not quite satisfied, as I cannot really distinguish syntax and semantics here and I'm not talking about neither structures nor evaluations, is the following:

(Right direction of the equivalence): We assume $\Delta,\psi\vDash\varphi$. Since $\varphi\rightarrow(\psi\rightarrow\varphi)$ is a tautology and we already know $\varphi$ to be valid from our assumption, we know that also $\psi\rightarrow\varphi$ is valid.

(Left direction): We assume $\Delta\vDash\psi\rightarrow\varphi$. Then we know also $\Delta,\psi\vDash\psi\rightarrow\varphi$ and hence $\Delta,\psi\vDash\varphi$.

Josef K.
  • 13

1 Answers1

0

Now I'm satisfied with the following reasoning:

(st) is met by (1) and my example of a structure for which there is one element a in P and one element b that is not in P is not a counterexample. This is so because in (st) we are not claiming that $\forall y P(y)$ is true for all evaluations in which $P(x)$ is true (which would be (cn) and is false), but something weaker: If $P(x)$ is true for all evaluations, then also $\forall y P(y)$ is true for all evaluations. But in this case the premise "P(x) is true for all evaluations" is false and therefore the implication true. So (st) holds.

As for the second part of my question:

We want to show $$\Delta,\varphi\models\psi\Leftrightarrow\Delta\models\varphi\rightarrow\psi$$

So for the right direction: Let e satisfy all formulas in $\Delta$. we want to show $\Delta\models(\varphi\rightarrow\psi)[e]$. Now there are two cases: 1. $\neg\varphi[e]$ and we are finished since then $\varphi[e]\rightarrow\psi$ is valid in any structure for any formula $\psi$ under any evaluation. 2. $\varphi[e]$ and then we know by the assumption of $\Delta,\varphi\models\psi$ that also $\psi[e]$ must be valid and hence $\Delta\models(\varphi\rightarrow\psi)[e]$.

For the left direction: Let e satisfy all formulas in $\Delta,\varphi$. We want to show: $\Delta,\varphi\models\psi[e]$. Now we know by assumption that $\Delta\models(\varphi\rightarrow\psi)[e]$. Since e also satisfies $\varphi$ we know that e must satisfy $\psi$ and hence $\Delta,\varphi\models\psi[e]$.

Josef K.
  • 13