reserve n,m for Nat,
  r,r1,r2,s,t for Real,
  x,y for set;

theorem Th72:
  for D be non empty set, F be PartFunc of D,REAL, r be Real, X be
set, d be Element of D st dom(F|X) is finite & d in dom(F|X) holds FinS(F-r,X).
  (len FinS(F-r,X)) = (F-r).d iff FinS(F,X).(len FinS(F,X)) = F.d
proof
  let D be non empty set, F be PartFunc of D,REAL, r be Real, X be set, d be
  Element of D;
  set dx = dom(F|X), drx = dom((F-r)|X), frx = FinS(F-r,X), fx = FinS(F,X);
  assume that
A1: dx is finite and
A2: d in dx;
  reconsider dx as finite set by A1;
A3: drx = dom(F-r) /\ X by RELAT_1:61
    .= dom F /\ X by VALUED_1:3
    .= dx by RELAT_1:61;
  then fx, F|X are_fiberwise_equipotent by Def13;
  then
A4: rng fx = rng(F|X) by CLASSES1:75;
  then fx <> {} by A2,FUNCT_1:3,RELAT_1:38;
  then 0+1<=len fx by NAT_1:13;
  then
A5: len fx in dom fx by FINSEQ_3:25;
  (F|X).d in rng(F|X) by A2,FUNCT_1:def 3;
  then F.d in rng(F|X) by A2,FUNCT_1:47;
  then consider n being Nat such that
A6: n in dom fx and
A7: fx.n = F.d by A4,FINSEQ_2:10;
A8: dom fx = Seg len fx by FINSEQ_1:def 3;
  frx, (F-r)|X are_fiberwise_equipotent by A3,Def13;
  then
A9: rng frx = rng((F-r)|X) by CLASSES1:75;
A10: len fx = card dx & dom frx = Seg len frx by Th67,FINSEQ_1:def 3;
A11: len frx = card dx by A3,Th67;
  then frx.(len frx) in rng frx by A10,A8,A5,FUNCT_1:def 3;
  then consider d1 be Element of D such that
A12: d1 in drx and
A13: ((F-r)|X).d1 = frx.(len frx) by A9,PARTFUN1:3;
  (F|X).d1 = F.d1 by A3,A12,FUNCT_1:47;
  then F.d1 in rng(F|X) by A3,A12,FUNCT_1:def 3;
  then consider m being Nat such that
A14: m in dom fx and
A15: fx.m = F.d1 by A4,FINSEQ_2:10;
A16: dom(F-r) = dom F by VALUED_1:3;
A17: drx = dom(F-r) /\ X by RELAT_1:61;
  then
A18: d1 in dom(F-r) by A12,XBOOLE_0:def 4;
A19: frx.(len frx) = (F-r).d1 by A12,A13,FUNCT_1:47
    .= F.d1 - r by A16,A18,VALUED_1:3;
A20: d in dom(F-r) by A2,A3,A17,XBOOLE_0:def 4;
  then
A21: (F-r).d = F.d - r by A16,VALUED_1:3;
A22: n<=len fx by A6,FINSEQ_3:25;
  thus frx.(len frx) = (F-r).d implies fx.(len fx) = F.d
  proof
    fx.(len fx) in rng fx by A5,FUNCT_1:def 3;
    then consider d1 be Element of D such that
A23: d1 in dx and
A24: (F|X).d1 = fx.(len fx) by A4,PARTFUN1:3;
A25: d1 in dom(F-r) by A3,A17,A23,XBOOLE_0:def 4;
A26: F.d1 = fx.(len fx) by A23,A24,FUNCT_1:47;
    ((F-r)|X).d1 = (F-r).d1 by A3,A23,FUNCT_1:47
      .= F.d1 - r by A16,A25,VALUED_1:3;
    then F.d1-r in rng((F-r)|X) by A3,A23,FUNCT_1:def 3;
    then consider m being Nat such that
A27: m in dom frx and
A28: frx.m = F.d1-r by A9,FINSEQ_2:10;
A29: m<=len frx by A27,FINSEQ_3:25;
    assume that
A30: frx.(len frx) = (F-r).d and
A31: fx.(len fx) <> F.d;
    n<len fx by A7,A22,A31,XXREAL_0:1;
    then
A32: F.d>=F.d1 by A5,A6,A7,A26,RFINSEQ:19;
    now
      per cases;
      case
        len frx = m;
        then F.d1+-r=F.d-r by A16,A20,A30,A28,VALUED_1:3;
        hence contradiction by A31,A23,A24,FUNCT_1:47;
      end;
      case
        len frx <>m;
        then m<len frx by A29,XXREAL_0:1;
        then F.d1-r>=F.d-r by A11,A10,A8,A21,A5,A30,A27,A28,RFINSEQ:19;
        then F.d1>=F.d by XREAL_1:9;
        hence contradiction by A31,A26,A32,XXREAL_0:1;
      end;
    end;
    hence contradiction;
  end;
  assume that
A33: fx.(len fx) = F.d and
A34: frx.(len frx) <> (F-r).d;
  ((F-r)|X).d in rng((F-r)|X) by A2,A3,FUNCT_1:def 3;
  then (F-r).d in rng((F-r)|X) by A2,A3,FUNCT_1:47;
  then consider n1 be Nat such that
A35: n1 in dom frx and
A36: frx.n1 = F.d -r by A9,A21,FINSEQ_2:10;
  n1<=len frx by A35,FINSEQ_3:25;
  then n1<len frx by A21,A34,A36,XXREAL_0:1;
  then F.d-r>=F.d1-r by A11,A10,A8,A5,A19,A35,A36,RFINSEQ:19;
  then
A37: F.d>=F.d1 by XREAL_1:9;
A38: m<=len fx by A14,FINSEQ_3:25;
  now
    per cases;
    case
      len fx = m;
      hence contradiction by A16,A20,A33,A34,A19,A15,VALUED_1:3;
    end;
    case
      len fx <> m;
      then m<len fx by A38,XXREAL_0:1;
      then F.d1>=F.d by A5,A33,A14,A15,RFINSEQ:19;
      hence contradiction by A21,A34,A19,A37,XXREAL_0:1;
    end;
  end;
  hence contradiction;
end;
