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Remark. Crossposted from Math SEMath SE due to lack of responses.

The following text appears in Janusz Czelakowski's "Protoalgebraic Logics":

Suppose there exists a class $\mathbf{K}$ of matrices for a language $S$ and a $k$-parameterized set $E(p,q,\underline{r})$ of sentence in $S$ such that for every matrix $M=(A, D)$ of $\mathbf{K}$ and every pair $a,b\in A$,

(a) $E(a, a, \underline{c})\subseteq D$ for all strings $\underline{c}\in A^k$;

(b) if $E(a,b,\underline{c})\subseteq D$ for all sequences $\underline{c}\in A^k$, then $a=b$.

Then the consequence relation $\mathbf{K}^\vDash$ in $S$ is protoalgebraic and $E(p,q,\underline{r})$ is a parameterized equivalence for $\mathbf{K}^\vDash$.

Maybe I missed something obvious, but I am not able to convince myself that this is true.

I tried to show protoalgebraicity by showing that $E(p,q,\underline{r})$ satisfies p-(R), p-(MP) and p-(RP) (paramaterized reflexivity, modus ponens and replacement respectively).

In the case of modus ponens, we need to show $q\in \mathbf{K}^\vDash(p, E(\langle p,q\rangle)$, in other words, for every matrix $(A, D)$ in $\mathbf{K}$ and every valuation $h:S\to A$, if $hp\in D$ and $hE(\langle p,q\rangle)\subseteq D$ then $hq\in D$. But $$h(E\langle p,q\rangle)=\bigcup \{E(hp, hq, h\underline{\gamma})\mid \underline{\gamma}\in S^k\},$$

and since $h$ need not be surjective, it's not a priori clear (and I don't believe it's true) that our set is equal to $$\bigcup \{E(hp, hq, \underline{c})\mid \underline{c}\in A^k\},$$ which is what one could use to conclude $hq=hp\in D$.

Remark. Crossposted from Math SE due to lack of responses.

The following text appears in Janusz Czelakowski's "Protoalgebraic Logics":

Suppose there exists a class $\mathbf{K}$ of matrices for a language $S$ and a $k$-parameterized set $E(p,q,\underline{r})$ of sentence in $S$ such that for every matrix $M=(A, D)$ of $\mathbf{K}$ and every pair $a,b\in A$,

(a) $E(a, a, \underline{c})\subseteq D$ for all strings $\underline{c}\in A^k$;

(b) if $E(a,b,\underline{c})\subseteq D$ for all sequences $\underline{c}\in A^k$, then $a=b$.

Then the consequence relation $\mathbf{K}^\vDash$ in $S$ is protoalgebraic and $E(p,q,\underline{r})$ is a parameterized equivalence for $\mathbf{K}^\vDash$.

Maybe I missed something obvious, but I am not able to convince myself that this is true.

I tried to show protoalgebraicity by showing that $E(p,q,\underline{r})$ satisfies p-(R), p-(MP) and p-(RP) (paramaterized reflexivity, modus ponens and replacement respectively).

In the case of modus ponens, we need to show $q\in \mathbf{K}^\vDash(p, E(\langle p,q\rangle)$, in other words, for every matrix $(A, D)$ in $\mathbf{K}$ and every valuation $h:S\to A$, if $hp\in D$ and $hE(\langle p,q\rangle)\subseteq D$ then $hq\in D$. But $$h(E\langle p,q\rangle)=\bigcup \{E(hp, hq, h\underline{\gamma})\mid \underline{\gamma}\in S^k\},$$

and since $h$ need not be surjective, it's not a priori clear (and I don't believe it's true) that our set is equal to $$\bigcup \{E(hp, hq, \underline{c})\mid \underline{c}\in A^k\},$$ which is what one could use to conclude $hq=hp\in D$.

Remark. Crossposted from Math SE due to lack of responses.

The following text appears in Janusz Czelakowski's "Protoalgebraic Logics":

Suppose there exists a class $\mathbf{K}$ of matrices for a language $S$ and a $k$-parameterized set $E(p,q,\underline{r})$ of sentence in $S$ such that for every matrix $M=(A, D)$ of $\mathbf{K}$ and every pair $a,b\in A$,

(a) $E(a, a, \underline{c})\subseteq D$ for all strings $\underline{c}\in A^k$;

(b) if $E(a,b,\underline{c})\subseteq D$ for all sequences $\underline{c}\in A^k$, then $a=b$.

Then the consequence relation $\mathbf{K}^\vDash$ in $S$ is protoalgebraic and $E(p,q,\underline{r})$ is a parameterized equivalence for $\mathbf{K}^\vDash$.

Maybe I missed something obvious, but I am not able to convince myself that this is true.

I tried to show protoalgebraicity by showing that $E(p,q,\underline{r})$ satisfies p-(R), p-(MP) and p-(RP) (paramaterized reflexivity, modus ponens and replacement respectively).

In the case of modus ponens, we need to show $q\in \mathbf{K}^\vDash(p, E(\langle p,q\rangle)$, in other words, for every matrix $(A, D)$ in $\mathbf{K}$ and every valuation $h:S\to A$, if $hp\in D$ and $hE(\langle p,q\rangle)\subseteq D$ then $hq\in D$. But $$h(E\langle p,q\rangle)=\bigcup \{E(hp, hq, h\underline{\gamma})\mid \underline{\gamma}\in S^k\},$$

and since $h$ need not be surjective, it's not a priori clear (and I don't believe it's true) that our set is equal to $$\bigcup \{E(hp, hq, \underline{c})\mid \underline{c}\in A^k\},$$ which is what one could use to conclude $hq=hp\in D$.

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Why does the following test for protoalgebraicity work?

Remark. Crossposted from Math SE due to lack of responses.

The following text appears in Janusz Czelakowski's "Protoalgebraic Logics":

Suppose there exists a class $\mathbf{K}$ of matrices for a language $S$ and a $k$-parameterized set $E(p,q,\underline{r})$ of sentence in $S$ such that for every matrix $M=(A, D)$ of $\mathbf{K}$ and every pair $a,b\in A$,

(a) $E(a, a, \underline{c})\subseteq D$ for all strings $\underline{c}\in A^k$;

(b) if $E(a,b,\underline{c})\subseteq D$ for all sequences $\underline{c}\in A^k$, then $a=b$.

Then the consequence relation $\mathbf{K}^\vDash$ in $S$ is protoalgebraic and $E(p,q,\underline{r})$ is a parameterized equivalence for $\mathbf{K}^\vDash$.

Maybe I missed something obvious, but I am not able to convince myself that this is true.

I tried to show protoalgebraicity by showing that $E(p,q,\underline{r})$ satisfies p-(R), p-(MP) and p-(RP) (paramaterized reflexivity, modus ponens and replacement respectively).

In the case of modus ponens, we need to show $q\in \mathbf{K}^\vDash(p, E(\langle p,q\rangle)$, in other words, for every matrix $(A, D)$ in $\mathbf{K}$ and every valuation $h:S\to A$, if $hp\in D$ and $hE(\langle p,q\rangle)\subseteq D$ then $hq\in D$. But $$h(E\langle p,q\rangle)=\bigcup \{E(hp, hq, h\underline{\gamma})\mid \underline{\gamma}\in S^k\},$$

and since $h$ need not be surjective, it's not a priori clear (and I don't believe it's true) that our set is equal to $$\bigcup \{E(hp, hq, \underline{c})\mid \underline{c}\in A^k\},$$ which is what one could use to conclude $hq=hp\in D$.