In (4), if we let ##\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(c, \lambda)\}## reduce to ##A(b, \lambda)A(c, \lambda)## by allowing ##A(a, \lambda)A(a, \lambda)=1,## (as stated) then the integral gives ##-E(b,c) ##. It is similar to the way you get the same final result.
You and I and Bell...