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Let $\mathcal{L}=\{e,\cdot\}$ be the monoids' language. Give a syntactic deduction about: $$\forall x (x \cdot e = x \land e \cdot x = x) \vdash \forall z (\forall x \hspace{2mm} z \cdot x = x) \rightarrow z=e$$

I am stuck with this, I tried the following:

\begin{matrix} \forall x (x \cdot e = x \land e \cdot x = x), \forall x \hspace{2mm} y \cdot x = x, \neg y=e \vdash \bot \\ \forall x (x \cdot e = x \land e \cdot x = x), \forall x \hspace{2mm} y \cdot x = x \vdash \neg \neg y=e & (\neg I) \\ \forall x (x \cdot e = x \land e \cdot x = x), \forall x \hspace{2mm} y \cdot x = x \vdash y=e & (\neg E)\\ \forall x (x \cdot e = x \land e \cdot x = x) \vdash (\forall x \hspace{2mm} y \cdot x = x) \rightarrow y=e & (\rightarrow I) \\ \forall x (x \cdot e = x \land e \cdot x = x) \vdash \forall z (\forall x \hspace{2mm} z \cdot x = x) \rightarrow z=e & (\forall I) \end{matrix}

but I don't see how to continue on. Possible ideas would be appreciated.

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  • $\begingroup$ It sounds like your previous post... $\endgroup$
    – Mauro ALLEGRANZA
    Jun 1 at 16:12
  • $\begingroup$ @MauroALLEGRANZA it is not the same formula because there is one more variable and the brackets are not in the same place. In fact, this formula and the one from the other post don't affirm the same $\endgroup$
    – Superdivinidad
    Jun 1 at 16:13

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