reserve f,g,h for Function,
  A for set;
reserve F for Function,
  B,x,y,y1,y2,z for set;
reserve x,z for object;

theorem Th23:
  f|A = g|A implies (F.:(f,h))|A = (F.:(g,h))|A
proof
  assume
A1: f|A = g|A;
  thus (F.:(f,h))|A = F*<:f,h:>|A by RELAT_1:83
    .= F*<:f|A,h:> by Th5
    .= F*<:g,h:>|A by A1,Th5
    .= (F.:(g,h))|A by RELAT_1:83;
end;
