theorem Th47:
  for A being disjoint_valued non-empty MSAlgebra over S
  for B being non-empty MSAlgebra over S
  for o being OperSymbol of S, p,q being Element of Args(o,A)
  for h being ManySortedFunction of A,B, a being Element of A
  for i st i in dom p & q = p+*(i,a) holds h#q = (h#p)+*(i,h.a)
  proof
    let A be disjoint_valued non-empty MSAlgebra over S;
    let B be non-empty MSAlgebra over S;
    let o be OperSymbol of S;
    let p,q be Element of Args(o,A);
    let h be ManySortedFunction of A,B;
    let a be Element of A;
    let i;
    assume Z0: i in dom p;
    assume Z1: q = p+*(i,a);
A1: dom (h#q) = dom the_arity_of o = dom (h#p) = dom((h#p)+*(i,h.a)) &
    dom q = dom the_arity_of o = dom p by MSUALG_3:6,FUNCT_7:30;
    now let j be object;
      assume
A2:   j in dom (h#q);
      then reconsider k = j as Nat;
A3:   (h#q).k = h.((the_arity_of o)/.k).(q.k) &
      (h#p).k = h.((the_arity_of o)/.k).(p.k) by A1,A2,MSUALG_3:def 6;
      q.i = a by Z0,Z1,FUNCT_7:31;
      then a in (the Sorts of A).((the_arity_of o)/.i) by Z0,A1,MSUALG_6:2;
      then the_sort_of a = (the_arity_of o)/.i by SORT;
      then
A5:   h.((the_arity_of o)/.i).a = h.a by ABBR;
      per cases;
      suppose
A4:     j = i;
        hence (h#q).j = h.((the_arity_of o)/.k).a by Z1,A1,A2,A3,FUNCT_7:31
        .= ((h#p)+*(i,h.a)).j by A4,A5,A1,A2,FUNCT_7:31;
      end;
      suppose
A6:     j <> i;
        hence (h#q).j = h.((the_arity_of o)/.k).(p.k) by Z1,A3,FUNCT_7:32
        .= ((h#p)+*(i,h.a)).j by A3,A6,FUNCT_7:32;
      end;
    end;
    hence h#q = (h#p)+*(i,h.a) by A1,FUNCT_1:2;
  end;
