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If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of JonsonJónsson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known?

Edit: as François G. Dorais pointed out in a comment, if $\kappa$ is Jónsson then $V_\kappa\models$ "for every $x$, $x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists".

If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known?

Edit: as François G. Dorais pointed out in a comment, if $\kappa$ is Jónsson then $V_\kappa\models$ "for every $x$, $x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists".

If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jónsson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known?

Edit: as François G. Dorais pointed out in a comment, if $\kappa$ is Jónsson then $V_\kappa\models$ "for every $x$, $x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists".

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If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known? Is it precisely equiconsistent with

Edit: as François G. Dorais pointed out in a proper class ofcomment, if $\kappa$ is Jónsson cardinalsthen $V_\kappa\models$ "for every $x$, or weaker?$x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists".

If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known? Is it precisely equiconsistent with a proper class of Jónsson cardinals, or weaker?

If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson cardinals is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known?

Edit: as François G. Dorais pointed out in a comment, if $\kappa$ is Jónsson then $V_\kappa\models$ "for every $x$, $x^\#$ exists". (we aren't guarenteed that $V_\kappa\models ZFC$ but that doesn't matter consistency-strength-wise), so "there is a Jónsson cardinal" is an upper bound for the consistency strength of "for every $x$, $x^\#$ exists".

added 13 characters in body
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If there exists a JonsonJónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of JonsonJónsson is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known? Is it precisely equiconsistent with a proper class of JonsonJónsson cardinals, or weaker?

If there exists a Jonson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jonson is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known? Is it precisely equiconsistent with a proper class of Jonson, or weaker?

If there exists a Jónsson cardinal $\kappa$, then $x^\#$ exists for every $x\in V_\kappa$ (in particular $V\neq L[x]$). It follows that if there is a proper class of Jonson cardinals, then the sharp of every set should exist (this happens if for example if there is a proper class of Ramsey or measurable cardinals).

So the existence of a proper class of Jónsson is an upper bound for the consistency of "for every $x$, $x^\#$ exists". $\mathbf{\Pi}^1_1$-determinacy, being equivalent to the existence of the sharp of every real, is a lower bound.

Is the exact consistency strength of "for every $x$, $x^\#$ exists" known? Is it precisely equiconsistent with a proper class of Jónsson cardinals, or weaker?

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