revise df-prjsp

[icecream17]

Restricts the domain of the equivalence relation to not include 0, to make theorems using the equivalence operator nicer

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sidenote: I'm stumped on trying to prove something like:

  ${
    $d ph y $.  $d t y $.  $d x y $.
    $( Equivalence between ~ df-ssb and ~ df-sb . $)
    sb11 $p |- ( A. y ( y = t -> A. x ( x = y -> ph ) ) <->
                 ( ( x = t -> ph ) /\ E. x ( x = t /\ ph ) ) ) $=
      ? $.
  $}

The reverse direction becomes ( x = t -> ph ) /\ y = t /\ E. x ( x = t /\ ph ) -> A. x ( x = y -> ph ) after a simple alrimiv which seems so possible but I can't find a solution??? ( x = y -> ph ) is easily provable but there's no easy way to add the quantifier. edit: invalid alrimiv because of $d v condition

[benjub]

sidenote: I'm stumped on trying to prove something like:

  ${
    $d ph y $.  $d t y $.  $d x y $.
    $( Equivalence between ~ df-ssb and ~ df-sb . $)
    sb11 $p |- ( A. y ( y = t -> A. x ( x = y -> ph ) ) <->
                 ( ( x = t -> ph ) /\ E. x ( x = t /\ ph ) ) ) $=
      ? $.
  $}

This is basically https://us.metamath.org/mpeuni/bj-dfssb2.html (its RHS is df-sb by https://us.metamath.org/mpeuni/dfsb7.html).

[tirix]

Looks good!

[tirix]

~f1oun seems to be the best top-level theorem to start with.
Then you can probably explicitly construct the mappings.

[wlammen]

@icecream17 left a comment in #3219 , so I think he saw the remarks here, too, and it is safe to merge this pull request.