
theorem Th11:
  for L being non empty multMagma, a being Element of L, p, q
  being FinSequence of the carrier of L holds (p^q)*a = (p*a)^(q*a)
proof
  let L be non empty multMagma, a be Element of L, p, q be FinSequence of the
  carrier of L;
A1: dom ((p^q)*a) = dom (p^q) by Def2;
A2: dom (q*a) = dom q by Def2;
  then
A3: len (q*a) = len q by FINSEQ_3:29;
A4: dom (p*a) = dom p by Def2;
  then
A5: len (p*a) = len p by FINSEQ_3:29;
A6: now
    let i be Nat;
    assume
A7: i in dom ((p^q)*a);
    per cases by A1,A7,FINSEQ_1:25;
    suppose
A8:   i in dom p;
      thus ((p^q)*a)/.i = ((p^q)/.i)*a by A1,A7,Def2
        .= (p/.i)*a by A8,FINSEQ_4:68
        .= (p*a)/.i by A8,Def2
        .= ((p*a)^(q*a))/.i by A4,A8,FINSEQ_4:68;
    end;
    suppose
      ex n being Nat st n in dom q & i = len p+n;
      then consider n being Nat such that
A9:   n in dom q and
A10:  i = len p+n;
      thus (p^q*a)/.i = ((p^q)/.i)*a by A1,A7,Def2
        .= (q/.n)*a by A9,A10,FINSEQ_4:69
        .= (q*a)/.n by A9,Def2
        .= ((p*a)^(q*a))/.i by A5,A2,A9,A10,FINSEQ_4:69;
    end;
  end;
  len ((p*a)^(q*a)) = len (p*a) + len (q*a) by FINSEQ_1:22
    .= len (p^q) by A5,A3,FINSEQ_1:22;
  then dom ((p^q)*a) = dom ((p*a)^(q*a)) by A1,FINSEQ_3:29;
  hence thesis by A6,FINSEQ_5:12;
end;
