theorem Th31:
  x in dom f & g = f.x & y in dom g implies [x,y] in dom uncurry f
  & (uncurry f).(x,y) = g.y & g.y in rng uncurry f
proof
  assume that
A1: x in dom f and
A2: g = f.x and
A3: y in dom g;
  thus
A4: [x,y] in dom uncurry f by A1,A2,A3,Def2;
  [x,y]`1 = x & [x,y]`2 = y;
  hence (uncurry f).(x,y) = g.y by A2,A4,Def2;
  hence thesis by A4,FUNCT_1:def 3;
end;
