theorem
  f <#> -g = <-> (f <#> g)
proof
  set f1 = f<#>g;
A1: dom(<->f1) = dom f1 by Def33;
  dom f1 = dom f /\ dom g & dom(f<#>-g) = dom f /\ dom -g by Def43;
  hence
A2: dom(f<#>-g) = dom(<->f1) by A1,VALUED_1:8;
  let x be object;
  assume
A3: x in dom(f<#>-g);
  hence (f<#>-g).x = f.x (#) (-g).x by Def43
    .= f.x(#)-g.x by VALUED_1:8
    .= -(f.x(#)g.x) by Th24
    .= -f1.x by A1,A2,A3,Def43
    .= (<->f1).x by A2,A3,Def33;
end;
