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Do You Have a Separate Inerpo Argument?
It's not that you're computing am an inner product argument within the recursive proof. It's that you're folding together am two arguments, a, in such a way that if the import arguments were valid, then the output argument will be valid too. So it'sou you do have this reduction step from two to one, but it's not thatYou're reducing the two iner p documents, of the mo inerpo arguments from the profs you're checking into this new one. You're just checking them in a way that means you have to produce a proof that itself has't a separate inner product argument that it generates. And i think we can say there was no trusted