Article FLANG_2, MML version 4.99.1005

:: FLANG_2:th 1
theorem
for b1, b2, b3, b4 being natural set
      st b1 + b2 <= b3 & b3 <= b4 + b2
   holds ex b5 being natural set st
      b5 + b2 = b3 & b1 <= b5 & b5 <= b4;

:: FLANG_2:th 2
theorem
for b1, b2, b3, b4, b5 being natural set
      st b1 <= b2 & b3 <= b4 & b1 + b3 <= b5 & b5 <= b2 + b4
   holds ex b6, b7 being natural set st
      b6 + b7 = b5 & b1 <= b6 & b6 <= b2 & b3 <= b7 & b7 <= b4;

:: FLANG_2:th 3
theorem
for b1, b2 being natural set
      st b1 < b2
   holds ex b3 being natural set st
      b1 + b3 = b2 & 0 < b3;

:: FLANG_2:th 4
theorem
for b1 being set
for b2, b3 being Element of b1 ^omega
      st (b2 ^ b3 = b2 or b3 ^ b2 = b2)
   holds b3 = {};

:: FLANG_2:th 5
theorem
for b1, b2 being set
for b3, b4 being Element of bool (b2 ^omega)
      st (b1 in b3 or b1 in b4) & b1 <> <%> b2
   holds b3 ^^ b4 <> {<%> b2};

:: FLANG_2:th 6
theorem
for b1, b2 being set
for b3, b4 being Element of bool (b2 ^omega) holds
   <%b1%> in b3 ^^ b4
iff
   (<%> b2 in b3 & <%b1%> in b4 or <%b1%> in b3 & <%> b2 in b4);

:: FLANG_2:th 7
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
for b4 being natural set
      st b1 in b3 & b1 <> <%> b2 & 0 < b4
   holds b3 |^ b4 <> {<%> b2};

:: FLANG_2:th 8
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
      <%> b1 in b2 |^ b3
   iff
      (b3 = 0 or <%> b1 in b2);

:: FLANG_2:th 9
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
for b4 being natural set holds
      <%b1%> in b3 |^ b4
   iff
      <%b1%> in b3 &
       (<%> b2 in b3 & 1 < b4 or b4 = 1);

:: FLANG_2:th 10
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
for b4, b5 being natural set
      st b4 <> b5 & b3 |^ b4 = {b1} & b3 |^ b5 = {b1}
   holds b1 = <%> b2;

:: FLANG_2:th 11
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
(b2 |^ b3) |^ b4 = (b2 |^ b4) |^ b3;

:: FLANG_2:th 12
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
(b2 |^ b3) ^^ (b2 |^ b4) = (b2 |^ b4) ^^ (b2 |^ b3);

:: FLANG_2:th 13
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4 being natural set
      st <%> b1 in b2
   holds b3 c= b3 ^^ (b2 |^ b4) & b3 c= (b2 |^ b4) ^^ b3;

:: FLANG_2:th 14
theorem
for b1 being set
for b2, b3, b4 being Element of bool (b1 ^omega)
for b5, b6 being natural set
      st b2 c= b3 |^ b5 & b4 c= b3 |^ b6
   holds b2 ^^ b4 c= b3 |^ (b5 + b6);

:: FLANG_2:th 15
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
      st b1 in b3 & b1 <> <%> b2
   holds b3 * <> {<%> b2};

:: FLANG_2:th 16
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set
      st <%> b1 in b2 & 0 < b3
   holds (b2 |^ b3) * = b2 *;

:: FLANG_2:th 17
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set
      st <%> b1 in b2
   holds (b2 |^ b3) * = b2 * |^ b3;

:: FLANG_2:th 18
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega) holds
b2 c= b2 ^^ (b3 *) & b2 c= b3 * ^^ b2;

:: FLANG_2:funcnot 1 => FLANG_2:func 1
definition
  let a1 be set;
  let a2 be Element of bool (a1 ^omega);
  let a3, a4 be natural set;
  func A2 |^(A3,A4) -> Element of bool (a1 ^omega) equals
    union {b1 where b1 is Element of bool (a1 ^omega): ex b2 being natural set st
       a3 <= b2 & b2 <= a4 & b1 = a2 |^ b2};
end;

:: FLANG_2:def 1
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
b2 |^(b3,b4) = union {b5 where b5 is Element of bool (b1 ^omega): ex b6 being natural set st
   b3 <= b6 & b6 <= b4 & b5 = b2 |^ b6};

:: FLANG_2:th 19
theorem
for b1, b2 being set
for b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set holds
   b2 in b3 |^(b4,b5)
iff
   ex b6 being natural set st
      b4 <= b6 & b6 <= b5 & b2 in b3 |^ b6;

:: FLANG_2:th 20
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set
      st b3 <= b4 & b4 <= b5
   holds b2 |^ b4 c= b2 |^(b3,b5);

:: FLANG_2:th 21
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
   b2 |^(b3,b4) = {}
iff
   (b3 <= b4 implies 0 < b3 & b2 = {});

:: FLANG_2:th 22
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 |^(b3,b3) = b2 |^ b3;

:: FLANG_2:th 23
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5, b6 being natural set
      st b3 <= b4 & b5 <= b6
   holds b2 |^(b4,b5) c= b2 |^(b3,b6);

:: FLANG_2:th 24
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set
      st b3 <= b4 & b4 <= b5
   holds b2 |^(b3,b5) = (b2 |^(b3,b4)) \/ (b2 |^(b4,b5));

:: FLANG_2:th 25
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set
      st b3 <= b4 & b4 <= b5
   holds b2 |^(b3,b5) = (b2 |^(b3,b4)) \/ (b2 |^(b4 + 1,b5));

:: FLANG_2:th 26
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4 + 1
   holds b2 |^(b3,b4 + 1) = (b2 |^(b3,b4)) \/ (b2 |^ (b4 + 1));

:: FLANG_2:th 27
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4
   holds b2 |^(b3,b4) = (b2 |^ b3) \/ (b2 |^(b3 + 1,b4));

:: FLANG_2:th 28
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 |^(b3,b3 + 1) = (b2 |^ b3) \/ (b2 |^ (b3 + 1));

:: FLANG_2:th 29
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set
      st b2 c= b3
   holds b2 |^(b4,b5) c= b3 |^(b4,b5);

:: FLANG_2:th 30
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
for b4, b5 being natural set
      st b1 in b3 & b1 <> <%> b2 & (b4 <= 0 implies 0 < b5)
   holds b3 |^(b4,b5) <> {<%> b2};

:: FLANG_2:th 31
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
   b2 |^(b3,b4) = {<%> b1}
iff
   ((b3 <= b4 implies b2 <> {<%> b1}) &
    (b3 = 0 implies b4 <> 0) implies b3 = 0 & b2 = {});

:: FLANG_2:th 32
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
b2 |^(b3,b4) c= b2 *;

:: FLANG_2:th 33
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
   <%> b1 in b2 |^(b3,b4)
iff
   (b3 = 0 or b3 <= b4 & <%> b1 in b2);

:: FLANG_2:th 34
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st <%> b1 in b2 & b3 <= b4
   holds b2 |^(b3,b4) = b2 |^ b4;

:: FLANG_2:th 35
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^(b3,b4)) ^^ (b2 |^ b5) = (b2 |^ b5) ^^ (b2 |^(b3,b4));

:: FLANG_2:th 36
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
(b2 |^(b3,b4)) ^^ b2 = b2 ^^ (b2 |^(b3,b4));

:: FLANG_2:th 37
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5, b6 being natural set
      st b3 <= b4 & b5 <= b6
   holds (b2 |^(b3,b4)) ^^ (b2 |^(b5,b6)) = b2 |^(b3 + b5,b4 + b6);

:: FLANG_2:th 38
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
b2 |^(b3 + 1,b4 + 1) = (b2 |^(b3,b4)) ^^ b2;

:: FLANG_2:th 39
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5, b6 being natural set holds
(b2 |^(b3,b4)) ^^ (b2 |^(b5,b6)) = (b2 |^(b5,b6)) ^^ (b2 |^(b3,b4));

:: FLANG_2:th 40
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^(b3,b4)) |^ b5 = b2 |^(b3 * b5,b4 * b5);

:: FLANG_2:th 41
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^ (b3 + 1)) |^(b4,b5) c= ((b2 |^ b3) |^(b4,b5)) ^^ (b2 |^(b4,b5));

:: FLANG_2:th 42
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^ b3) |^(b4,b5) c= b2 |^(b3 * b4,b3 * b5);

:: FLANG_2:th 43
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^ b3) |^(b4,b5) c= (b2 |^(b4,b5)) |^ b3;

:: FLANG_2:th 44
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5, b6 being natural set holds
(b2 |^ (b3 + b4)) |^(b5,b6) c= ((b2 |^ b3) |^(b5,b6)) ^^ ((b2 |^ b4) |^(b5,b6));

:: FLANG_2:th 45
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 |^(0,0) = {<%> b1};

:: FLANG_2:th 46
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 |^(0,1) = {<%> b1} \/ b2;

:: FLANG_2:th 47
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 |^(1,1) = b2;

:: FLANG_2:th 48
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 |^(0,2) = ({<%> b1} \/ b2) \/ (b2 ^^ b2);

:: FLANG_2:th 49
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 |^(1,2) = b2 \/ (b2 ^^ b2);

:: FLANG_2:th 50
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 |^(2,2) = b2 ^^ b2;

:: FLANG_2:th 51
theorem
for b1, b2 being set
for b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set
   st 0 < b4 & b4 <> b5 & b3 |^(b4,b5) = {b2}
for b6 being natural set
      st b4 <= b6 & b6 <= b5
   holds b3 |^ b6 = {b2};

:: FLANG_2:th 52
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
for b4, b5 being natural set
      st b4 <> b5 & b3 |^(b4,b5) = {b1}
   holds b1 = <%> b2;

:: FLANG_2:th 53
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
for b4, b5 being natural set holds
   <%b1%> in b3 |^(b4,b5)
iff
   <%b1%> in b3 &
    b4 <= b5 &
    (<%> b2 in b3 & 0 < b5 or b4 <= 1 & 1 <= b5);

:: FLANG_2:th 54
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set holds
(b2 /\ b3) |^(b4,b5) c= (b2 |^(b4,b5)) /\ (b3 |^(b4,b5));

:: FLANG_2:th 55
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set holds
(b2 |^(b4,b5)) \/ (b3 |^(b4,b5)) c= (b2 \/ b3) |^(b4,b5);

:: FLANG_2:th 56
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5, b6 being natural set holds
(b2 |^(b3,b4)) |^(b5,b6) c= b2 |^(b3 * b5,b4 * b6);

:: FLANG_2:th 57
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set
      st b4 <= b5 & <%> b1 in b2
   holds b3 c= b3 ^^ (b2 |^(b4,b5)) & b3 c= (b2 |^(b4,b5)) ^^ b3;

:: FLANG_2:th 58
theorem
for b1 being set
for b2, b3, b4 being Element of bool (b1 ^omega)
for b5, b6, b7, b8 being natural set
      st b5 <= b6 & b7 <= b8 & b2 c= b3 |^(b5,b6) & b4 c= b3 |^(b7,b8)
   holds b2 ^^ b4 c= b3 |^(b5 + b7,b6 + b8);

:: FLANG_2:th 59
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
(b2 |^(b3,b4)) * c= b2 *;

:: FLANG_2:th 60
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
b2 * |^(b3,b4) c= b2 *;

:: FLANG_2:th 61
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4 & 0 < b4
   holds b2 * |^(b3,b4) = b2 *;

:: FLANG_2:th 62
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4 & 0 < b4 & <%> b1 in b2
   holds (b2 |^(b3,b4)) * = b2 *;

:: FLANG_2:th 63
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4 & <%> b1 in b2
   holds (b2 |^(b3,b4)) * = b2 * |^(b3,b4);

:: FLANG_2:th 64
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set
      st b2 c= b3 *
   holds b2 |^(b4,b5) c= b3 *;

:: FLANG_2:th 65
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
for b4, b5 being natural set
      st b2 c= b3 *
   holds b3 * = (b3 \/ (b2 |^(b4,b5))) *;

:: FLANG_2:th 66
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
(b2 |^(b3,b4)) ^^ (b2 *) = b2 * ^^ (b2 |^(b3,b4));

:: FLANG_2:th 67
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st <%> b1 in b2 & b3 <= b4
   holds b2 * = b2 * ^^ (b2 |^(b3,b4));

:: FLANG_2:th 68
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^(b3,b4)) |^ b5 c= b2 *;

:: FLANG_2:th 69
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set holds
(b2 |^ b3) |^(b4,b5) c= b2 *;

:: FLANG_2:th 70
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4
   holds (b2 |^ b3) * c= (b2 |^(b3,b4)) *;

:: FLANG_2:th 71
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5, b6 being natural set holds
(b2 |^(b3,b4)) |^(b5,b6) c= b2 *;

:: FLANG_2:th 72
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4, b5 being natural set
      st <%> b1 in b2 & b3 <= b4 & b5 <= b4
   holds b2 |^(b3,b4) = b2 |^(b5,b4);

:: FLANG_2:funcnot 2 => FLANG_2:func 2
definition
  let a1 be set;
  let a2 be Element of bool (a1 ^omega);
  func A2 ? -> Element of bool (a1 ^omega) equals
    union {b1 where b1 is Element of bool (a1 ^omega): ex b2 being natural set st
       b2 <= 1 & b1 = a2 |^ b2};
end;

:: FLANG_2:def 2
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? = union {b3 where b3 is Element of bool (b1 ^omega): ex b4 being natural set st
      b4 <= 1 & b3 = b2 |^ b4};

:: FLANG_2:th 73
theorem
for b1, b2 being set
for b3 being Element of bool (b1 ^omega) holds
      b2 in b3 ?
   iff
      ex b4 being natural set st
         b4 <= 1 & b2 in b3 |^ b4;

:: FLANG_2:th 74
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set
      st b3 <= 1
   holds b2 |^ b3 c= b2 ?;

:: FLANG_2:th 75
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? = (b2 |^ 0) \/ (b2 |^ 1);

:: FLANG_2:th 76
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? = {<%> b1} \/ b2;

:: FLANG_2:th 77
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 c= b2 ?;

:: FLANG_2:th 78
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega) holds
      b1 in b3 ?
   iff
      (b1 = <%> b2 or b1 in b3);

:: FLANG_2:th 79
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? = b2 |^(0,1);

:: FLANG_2:th 80
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
      b2 ? = b2
   iff
      <%> b1 in b2;

:: FLANG_2:funcreg 1
registration
  let a1 be set;
  let a2 be Element of bool (a1 ^omega);
  cluster a2 ? -> non empty;
end;

:: FLANG_2:th 81
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? ? = b2 ?;

:: FLANG_2:th 82
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
      st b2 c= b3
   holds b2 ? c= b3 ?;

:: FLANG_2:th 83
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
      st b1 in b3 & b1 <> <%> b2
   holds b3 ? <> {<%> b2};

:: FLANG_2:th 84
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
      b2 ? = {<%> b1}
   iff
      (b2 = {} or b2 = {<%> b1});

:: FLANG_2:th 85
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 * ? = b2 * & b2 ? * = b2 *;

:: FLANG_2:th 86
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? c= b2 *;

:: FLANG_2:th 87
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega) holds
(b2 /\ b3) ? = b2 ? /\ (b3 ?);

:: FLANG_2:th 88
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega) holds
b2 ? \/ (b3 ?) = (b2 \/ b3) ?;

:: FLANG_2:th 89
theorem
for b1, b2 being set
for b3 being Element of bool (b2 ^omega)
      st b3 ? = {b1}
   holds b1 = <%> b2;

:: FLANG_2:th 90
theorem
for b1, b2 being set
for b3 being Element of bool (b1 ^omega) holds
      <%b2%> in b3 ?
   iff
      <%b2%> in b3;

:: FLANG_2:th 91
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? ^^ b2 = b2 ^^ (b2 ?);

:: FLANG_2:th 92
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? ^^ b2 = b2 |^(1,2);

:: FLANG_2:th 93
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? ^^ (b2 ?) = b2 |^(0,2);

:: FLANG_2:th 94
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 ? |^ b3 = b2 ? |^(0,b3);

:: FLANG_2:th 95
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 ? |^ b3 = b2 |^(0,b3);

:: FLANG_2:th 96
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4
   holds b2 ? |^(b3,b4) = b2 ? |^(0,b4);

:: FLANG_2:th 97
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 ? |^(0,b3) = b2 |^(0,b3);

:: FLANG_2:th 98
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set
      st b3 <= b4
   holds b2 ? |^(b3,b4) = b2 |^(0,b4);

:: FLANG_2:th 99
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   (b2 |^(1,b3)) ? = b2 |^(0,b3);

:: FLANG_2:th 100
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
      st <%> b1 in b2 & <%> b1 in b3
   holds b2 ? c= b2 ^^ b3 & b2 ? c= b3 ^^ b2;

:: FLANG_2:th 101
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega) holds
b2 c= b2 ^^ (b3 ?) & b2 c= b3 ? ^^ b2;

:: FLANG_2:th 102
theorem
for b1 being set
for b2, b3, b4 being Element of bool (b1 ^omega)
      st b2 c= b3 ? & b4 c= b3 ?
   holds b2 ^^ b4 c= b3 |^(0,2);

:: FLANG_2:th 103
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set
      st <%> b1 in b2 & 0 < b3
   holds b2 ? c= b2 |^ b3;

:: FLANG_2:th 104
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 ? ^^ (b2 |^ b3) = (b2 |^ b3) ^^ (b2 ?);

:: FLANG_2:th 105
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
      st b2 c= b3 *
   holds b2 ? c= b3 *;

:: FLANG_2:th 106
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
      st b2 c= b3 *
   holds b3 * = (b3 \/ (b2 ?)) *;

:: FLANG_2:th 107
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? ^^ (b2 *) = b2 * ^^ (b2 ?);

:: FLANG_2:th 108
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? ^^ (b2 *) = b2 *;

:: FLANG_2:th 109
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 ? |^ b3 c= b2 *;

:: FLANG_2:th 110
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   (b2 |^ b3) ? c= b2 *;

:: FLANG_2:th 111
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
b2 ? ^^ (b2 |^(b3,b4)) = (b2 |^(b3,b4)) ^^ (b2 ?);

:: FLANG_2:th 112
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3 being natural set holds
   b2 ? ^^ (b2 |^ b3) = b2 |^(b3,b3 + 1);

:: FLANG_2:th 113
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
b2 ? |^(b3,b4) c= b2 *;

:: FLANG_2:th 114
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega)
for b3, b4 being natural set holds
(b2 |^(b3,b4)) ? c= b2 *;

:: FLANG_2:th 115
theorem
for b1 being set
for b2 being Element of bool (b1 ^omega) holds
   b2 ? = (b2 \ {<%> b1}) ?;

:: FLANG_2:th 116
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
      st b2 c= b3 ?
   holds b2 ? c= b3 ?;

:: FLANG_2:th 117
theorem
for b1 being set
for b2, b3 being Element of bool (b1 ^omega)
      st b2 c= b3 ?
   holds b3 ? = (b3 \/ b2) ?;