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I don't understand how the semantic restrictions for $\forall-$introduction and $\exists-$elimination work. These are $\dfrac{\Gamma \vdash \phi}{\Gamma\vdash \forall x\phi}, (x\notin FV(\Gamma))\quad$ and $\dfrac{\Gamma\vdash \exists x \phi\;\;\; \Gamma,\phi\vdash \psi}{\Gamma \vdash \psi}(x\notin FV(\Gamma, \psi))$.

I am supposed to show that for $\phi \equiv \phi(x,y)\quad$ it is possible to show $\exists x \forall y \phi(x,y)\implies\forall y \exists x \phi(x,y)\qquad$ taking into account the restrictions above; and without considering the restriction it is possible to show $\forall y \exists x \phi(x,y)\implies\exists x \forall y \phi (x,y).$

Looks like we could show $$\exists x \forall y \phi(x,y)\implies\forall y \exists x \phi(x,y)$$

by means of

$$\dfrac{\Gamma\vdash \exists x \forall y \phi(x,y)\;\;\; \Gamma,\forall y \phi(x,y)\vdash \psi}{\Gamma \vdash \psi}(x\notin FV(\Gamma,\psi))$$

Senex Ægypti Parvi
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Cure
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1 Answers1

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$$ \newcommand{\DeductionBox} [1]{\begin{array} {l|} #1 \\ \hline \end{array}} $$

$$\begin{array} {l} \DeductionBox{ \begin{array} {rlr} \quad (1) & \exists \langle x,~ \forall \langle y,~ Mxy \rangle ~ \rangle \quad & \text{New Assumption} \\ \end{array} \\ \\ \DeductionBox{ \begin{array} {rlr} \quad \quad (2) & \forall \langle y,~ Mxy \rangle \quad & \text{New Assumption}\\ \quad \quad (3) & Mxy \quad & \forall-\text{Elimination of }(2)\\ \quad \quad (4) & \exists \langle x ,~ Mxy \rangle \quad & \exists-\text{Introduction of }(3)\\ \end{array} } \\ \\ \begin{array} {rlr} \quad (5) & \exists \langle x ,~ Mxy \rangle \quad & \exists-\text{Elimination of }(1),~(2) \vdash (4) \\ \quad (6) & \forall \langle y,~ \exists \langle x,~ Mxy \rangle ~ \rangle \quad & \forall-\text{Introduction of } (5) \\ \end{array} } \\ \\ \begin{array} {rlr} (7) & \exists \langle x,~ \forall \langle y,~ Mxy \rangle ~ \rangle \implies \forall \langle y,~ \exists \langle x,~ Mxy \rangle ~ \rangle \quad & \text{Deduction of } (1) \vdash (6) \\ \end{array} \\ \end{array}$$

Anyway, one tip for constructing this proof, notice that it is a theorem of the form $\text{Assumption} \implies \text{Conclusion}$ but $\text{Assumption} \iff \text{Conclusion}$ isn't true. This means that your conclusion loses information present in the assumption. Since $\forall-\text{Introduction}$ and $\forall-\text{Elimination}$ are equivalences (they don't lose information), you are going to have to throw the extra information from the assumption away using $\exists-\text{Introduction}$ and $\exists-\text{Elimination}$ inferences.

Or to put it another way, look at a model (or visualization) of the assumption (such as a lattice with a colored column) and a model (or visualization) of the conclusion (such as a lattice where each row has an element colored) and see which rules of inferences allow you to eliminate the existentially quantified "column" assumption.

As requested, showing the contrapositive by breaking the variable scoping conditions. The sophistry proceeds by showing that $Mxy$ itself can be constructed from a misuse of $\exists-\text{Elimination}$, and the theorem follows easily from that.

$$\begin{array} {l} \DeductionBox { \begin{array} {rlr} \quad (1) & \forall \langle y,~ \exists \langle x,~ Mxy \rangle ~ \rangle & \text{New Assumption} \\ \quad (2) & \exists \langle x,~ Mxy \rangle & \forall-\text{Elimination of }(1) \\ \end{array} \\ \\ \DeductionBox { \begin{array} {rlr} \quad \quad (3) & Mxy & \text{New Assumption} \\ \end{array} }\\ \\ \begin{array} {rlr} \quad (4) & Mxy & \exists-\text{Elimination of }(2), (3) \vdash (3) \\ \quad (5) & \forall \langle y,~ Mxy \rangle & \forall-\text{Introduction of }(4) \\ \quad (6) & \exists \langle x,~ \forall \langle y,~ Mxy \rangle ~ \rangle & \exists-\text{Introduction of }(5) \\ \end{array} } \\ \\ \begin{array} {rlr} (7) & \forall \langle y,~ \exists \langle x,~ Mxy \rangle ~ \rangle \implies \exists \langle x,~ \forall \langle y,~ Mxy \rangle ~ \rangle & \text{Deduction of }(1) \vdash (6) \\ \end{array} \end{array}$$

In the $\exists-\text{Elimination}$ of step (4), the free variable $x$ violates variable scope conditions.

Footnote, I don't really like the style of proof of the first theorem though, as it hides the "why" behind the theorem. I would prefer to label the coordinates in the assumption $(x_1, y_1)$, and the coordinates in the conclusion $(x_2, y_2)$. Then to proceed by proof by contradiction. Looking at $M(x_1, y_2)$ , it must be both true (due to the condition) and false (due to the negated conclusion).

DanielV
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