reserve x,y,t for Real;

theorem
  x^2<1 & y^2<1 implies tanh"(x)+tanh"(y)=tanh"((x+y)/(1+x*y))
proof
  assume
A1: x^2<1 & y^2<1;
  then
A2: 0<(1+x)/(1-x) & 0<(1+y)/(1-y) by Lm4;
A3: tanh"(x)+tanh"(y) =(1/2)*(log(number_e,(1+x)/(1-x))+log(number_e,(1+y)/(
  1-y)))
    .=(1/2)*(log(number_e,((1+x)/(1-x))*((1+y)/(1-y)))) by A2,Lm1,POWER:53
,TAYLOR_1:11
    .=(1/2)*(log(number_e,(((1+x)*(1+y))/((1-x)*(1-y))))) by XCMPLX_1:76
    .=(1/2)*(log(number_e,((1+x+y+x*y)/(1+x*y-x-y))));
A4: x*y+1<>0 by A1,Th27;
  then
  tanh"((x+y)/(1+x*y)) =(1/2)*log(number_e,(((x+y)+(1+x*y)*1)/(1+x*y))/(1-
  ((x+y)/(1+x*y)))) by XCMPLX_1:113
    .=(1/2)*log(number_e,((x+y+1+x*y)/(1+x*y))/(((1*(1+x*y)-(x+y))/(1+x*y)))
  ) by A4,XCMPLX_1:127
    .=(1/2)*log(number_e,(1+x+y+x*y)/(1+x*y-x-y)) by A4,XCMPLX_1:55;
  hence thesis by A3;
end;
