theorem Th56:
  p " A c= (p ^ q) " A
proof
  let x be object;
A1: dom p c= dom(p ^ q) by FINSEQ_1:26;
  assume
A2: x in p " A; then
A3: x in dom p by FUNCT_1:def 7;
  then reconsider k = x as Element of NAT;
A4: p.k in A by A2,FUNCT_1:def 7;
  p.k = (p ^ q).k by A3,FINSEQ_1:def 7;
  hence thesis by A3,A1,A4,FUNCT_1:def 7;
end;
