bblum

 64 proof qcurry : ((?x:t. B(x)) => C) => (!x:t. B(x) => C) = 65 begin 66   [((?x:t. B(x)) => C); 67     [!x:t. B(x); 68 69       C; 70     ]; 71     !x:t. B(x) => C; 72   ]; 73   ((?x:t. B(x)) => C) => (!x:t. B(x) => C); 74 end;