 reserve L for Quasi-Boolean_Algebra,
         x, y, z for Element of L;
 reserve L for Nelson_Algebra,
         a, b, c, d, x, y, z for Element of L;

theorem
  for L be non empty NelsonStr st
    L is satisfying_N0* holds
  L is Nelson_Algebra iff
    L is satisfying_N1* satisfying_N2* satisfying_N3* satisfying_N4*
    satisfying_N5* satisfying_N6* satisfying_N7* satisfying_N8*
    satisfying_N9* satisfying_N10* satisfying_N11* satisfying_N12*
    satisfying_N13* satisfying_N14* satisfying_N15* satisfying_N16*
    satisfying_N17* satisfying_N18* satisfying_N19*
  proof
    let L be non empty NelsonStr;
    assume
A1: L is satisfying_N0*;
    thus L is Nelson_Algebra implies L is satisfying_N1* satisfying_N2*
    satisfying_N3* satisfying_N4* satisfying_N5* satisfying_N6*
    satisfying_N7* satisfying_N8* satisfying_N9* satisfying_N10*
    satisfying_N11* satisfying_N12* satisfying_N13* satisfying_N14*
    satisfying_N15* satisfying_N16* satisfying_N17* satisfying_N18*
    satisfying_N19*;
    assume
A2:   L is satisfying_N1* satisfying_N2* satisfying_N3* satisfying_N4*
      satisfying_N5* satisfying_N6* satisfying_N7* satisfying_N8*
      satisfying_N9* satisfying_N10* satisfying_N11* satisfying_N12*
      satisfying_N13* satisfying_N14* satisfying_N15* satisfying_N16*
      satisfying_N17* satisfying_N18* satisfying_N19*; then
    reconsider L1 = L as DeMorgan non empty NelsonStr;
A3: for a, b being Element of L1 holds a "/\" b = Top L1 implies
      a = Top L1 & b = Top L1
    proof
      let a, b be Element of L1;
      assume a "/\" b = (Top L1); then
      Top L1 < a & Top L1 < b by A2;
      hence thesis by A2;
    end;
    set w = Top L1;
A4: w => w = w by A2; then
    w => ((w => w) => (w => (w "/\" w))) = w by A2; then
    (w => (w => (w "/\" w))) = w by A4,A2; then
    (w => (w "/\" w)) = w by A2; then
A5: w "/\" w = w by A2;
A6: for a, b being Element of L1 holds a <= b iff a < b & -b < -a
    proof
      let a, b be Element of L1;
A7:   a =-> b = (a => b) "/\" ((-b) => (-a)) by A2;
      thus a <= b implies a < b & -b < -a
      proof
        assume a <= b; then
        a =-> b = Top L1 by A1;
        hence thesis by A3,A7;
      end;
      assume a < b & -b < -a;
      hence thesis by A7,A5,A1;
    end;
    set d = (Top L)`;
A8: for a being Element of L holds a <= Top L
    proof
      let a be Element of L;
      a => Top L = (Top L) => (a => Top L) by A2; then
A9:   a < Top L by A2;
      (-Top L) => (Top L => -a) = Top L by A2; then
      -Top L < -a by A2;
      hence thesis by A6,A9;
    end;
A10: for a being Element of L holds d <= a
    proof
      let a be Element of L;
      (-Top L) => (Top L => a) = Top L by A2; then
A11:  -Top L < a by A2;
      -a <= Top L by A8; then
      -a < Top L by A2; then
      -a < -d by A2;
      hence thesis by A11,A6;
    end;
A12: for a being Element of L holds d "/\" a = d
    proof
      let a be Element of L;
      d <= a by A10;
      hence thesis;
    end;
A13: for a being Element of L1 holds a => (Top L1) = (Top L1)
    proof
      let a be Element of L1;
      (Top L1) => (a => (Top L1)) = Top L1 by A2;
      hence thesis by A2;
    end;
A14: for a, b, c being Element of L1 holds
      a => b = (Top L1) & b => c = (Top L1) implies a => c = (Top L1)
    proof
      let a, b, c be Element of L1;
      assume
A15:  (a => b) = (Top L1) & b => c = (Top L1);
      (a => c) = (Top L1) => (a => c) by A2
         .= (Top L1) => ((Top L1) => (a => c)) by A2
         .= (a => (Top L1)) => ((Top L1) => (a => c)) by A13
         .= (Top L1) by A2,A15;
      hence thesis;
    end;
A16: L1 is satisfying_A1b
    proof
      let a, b, c be Element of L1;
      assume a < b & b < c;
      hence thesis by A14;
    end;
A17: L1 is satisfying_N6
    proof
      let a, b, c be Element of L1;
      assume
A18:  a < c & b < c;
      (a => c) => ((b => c) => ((a "\/" b) => c)) = Top L1 by A2; then
      ((b => c) => ((a "\/" b) => c)) = Top L1 by A18,A2;
      hence thesis by A2,A18;
    end;
A19: for a being Element of L1 holds a => a = Top L1
    proof
      let a be Element of L1;
A20:  (a => ((a => a) => a)) => ((a => (a => a)) => (a => a)) = Top L1
         by A2;
      a => ((a => a) => a) = Top L1 by A2; then
A21:  ((a => (a => a)) => (a => a)) = Top L1 by A20,A2;
      a => (a => a) = Top L1 by A2;
      hence thesis by A21,A2;
    end;
A22: L1 is satisfying_N7
    proof
      let a, b, c be Element of L1;
      assume
A23:  a < b & a < c;
      (a => b) => ((a => c) => (a => (b "/\" c))) = Top L1 by A2; then
      (a => c) => (a => (b "/\" c)) = Top L1 by A23,A2;
      hence thesis by A2,A23;
    end;
A24: for a, b being Element of L1 holds b < a "\/" b by A2;
A25: for a, b being Element of L1 holds a "/\" b <= a
    proof
      let a, b be Element of L1;
A26:  -(a "/\" b) = (-a) "\/" (-b) by A2;
A27:  a "/\" b < a by A2;
      -a < (-a) "\/" (-b) by A2;
      hence thesis by A27,A6,A26;
   end;

A28: for a, b being Element of L1 holds a <= a "\/" b
   proof
     let a, b be Element of L1;
A29: (-a) "/\" (-b) < -a by A2;
A30: a < a "\/" b by A2;
     -(a "\/" b) = (-a) "/\" (-b) by A2;
     hence thesis by A29,A30,A6;
   end;

A31: for a, b being Element of L1 holds b <= a "\/" b
   proof
     let a, b be Element of L1;
A32: b < a "\/" b by A2;
A33: -(a "\/" b) = (-a) "/\" (-b) by A2;
     (-a) "/\" (-b) < -b by A2;
     hence thesis by A6,A32,A33;
   end;

A34: for a, b being Element of L1 holds a "/\" b <= b
   proof
     let a, b be Element of L1;
A35: -(a "/\" b) = (-a) "\/" (-b) by A2;
A36: a "/\" b < b by A2;
     -b < (-a) "\/" (-b) by A2;
     hence thesis by A35,A36,A6;
   end;

A37: for a being Element of L1 holds a =-> a = Top L1
   proof
     let a be Element of L1;
     a =-> a = (a => a) "/\" ((-a) => -a) by A2
            .= (Top L1) "/\" ((-a) => -a) by A19
            .= Top L1 by A19,A5;
     hence thesis;
   end;

A38:
  for a, b being Element of L1 holds
    a = b iff a =-> b = Top L1 & b =-> a = Top L1
   proof
     let a, b be Element of L1;
     a =-> b = Top L1 & b =-> a = Top L implies a = b
     proof
       assume
A39:   a =-> b = Top L1 & b =-> a = Top L; then
       (b =-> a) => ((a =-> b) => a) = (Top L1) => ((b =-> a) => b)
         by A2; then
       (b =-> a) => ((a =-> b) => a) = (Top L1) => b by A2,A39; then
       (b =-> a) => ((Top L1) => a) = b by A39,A2; then
       (Top L1) => a = b by A39,A2;
       hence thesis by A2;
     end;
     hence thesis by A37;
   end;

A40: for a, b being Element of L1 holds a <= b & b <= a iff a = b
   proof
     let a, b be Element of L1;
     thus a <= b & b <= a implies a = b
     proof
       assume a <= b & b <= a; then
       a =-> b = Top L & b =-> a = Top L by A1;
       hence thesis by A38;
     end;
     assume a = b; then
     a =-> b = Top L & b =-> a = Top L by A38;
     hence thesis by A1;
   end;

A41: for b being Element of L1 holds Top L1 < b implies b = Top L1 by A2;
A42: L1 is satisfying_A1
   proof
     let a be Element of L1;
     thus thesis by A19;
   end;
A43: for a, b, c being Element of L1 holds
      a < b implies b => c < a => c & c => a < c => b
   proof
     let a, b, c be Element of L1;
     assume
A44: a < b;
A45: b => c < a => (b => c) by A2;
     a => (b => c) < ((a => b) => (a => c)) by A2; then
A46: (b => c) < ((a => b) => (a => c)) by A45,A16;
     (c => (Top L1)) => ((c => a) => (c => b)) = (Top L1) by A2,A44; then
     (Top L1) => ((c => a) => (c => b)) = (Top L1) by A13;
     hence thesis by A46,A2,A44;
   end;

A47: for a, b being Element of L1 holds a => (b => (a "/\" b)) = Top L1
   proof
     let a, b be Element of L1;
     (b => a) < ((b => b) => (b => (a "/\" b))) by A2; then
     (b => a) < ((Top L1) => (b => (a "/\" b))) by A19; then
     (b => a) < (b => (a "/\" b)) by A2; then
     a => (b => a) < a => (b => (a "/\" b)) by A43; then
     (Top L1) < a => (b => (a "/\" b)) by A2;
     hence thesis by A2;
   end;

A48:  for a,b,c being Element of L1 st
      a < (b => c) holds b < (a => c)
   proof
     let a, b, c be Element of L1;
     assume
A49: a < (b => c);
     a => (b => c) < (a => b) => (a => c) by A2; then
A50: (a => b) < (a => c) by A2,A49;
     b < (a => b) by A2;
     hence thesis by A50,A16;
   end;

A51: for a, c being Element of L1 holds a => (a => c) < a => c
   proof
     let a, c be Element of L1;
     a => (a => c) < (a => a) => (a => c) by A2; then
     a => (a => c) < (Top L1) => (a => c) by A19;
     hence thesis by A2;
   end;

A52: L1 is satisfying_N3
   proof
     let x, a, b be Element of L1;
A53: a "/\" x < b implies x < a => b
     proof
       assume a "/\" x < b; then
       (x => (a "/\" x)) < (x => b) by A43; then
A54:   a => (x => (a "/\" x)) < a => (x => b) by A43;
       a => (x => (a "/\" x)) = (Top L1) by A47; then
       a < (x => b) by A2,A54;
       hence thesis by A48;
     end;
     x < a => b implies a "/\" x < b
     proof
       assume
A55:   x < a => b;
       (a "/\" x) < x by A2; then
       (a "/\" x) < a => b by A55,A16; then
A56:   a < (a "/\" x) => b by A48;
       a "/\" x < a by A2; then
       a "/\" x < (a "/\" x) => b by A16,A56;
       hence thesis by A41,A51;
     end;
     hence thesis by A53;
   end;
A57: for a, b, c being Element of L1 st
       b < c holds a "/\" b < a "/\" c
   proof
     let a, b, c be Element of L1;
     assume
A58: b < c;
     a "/\" b < b by A2; then
A59: a "/\" b < c by A58,A16;
     a "/\" b < a by A2;
     hence thesis by A22,A59;
   end;

A58: for a, b, c being Element of L1 st
      b < c holds a "\/" b < a "\/" c
   proof
     let a, b, c be Element of L1;
     assume
A60: b < c;
A61: a < a "\/" c by A2;
     c < a "\/" c by A2; then
     b < a "\/" c by A60,A16;
     hence thesis by A61,A17;
   end;

A62: for a, b, c being Element of L1 st
     a <= c & b <= c holds a "\/" b <= c
   proof
     let a, b, c be Element of L1;
     assume
A63: a <= c & b <= c; then
A64: a < c & -c < -a by A6;
A65: b < c & -c < -b by A63,A6;
     ((-c) => (-a)) => (((-c) => (-b)) => ((-c) => ((-a) "/\" -b))) = Top L1
       by A2; then
     (((-c) => (-b)) => ((-c) => ((-a) "/\" -b))) = Top L1 by A64,A2; then
     ((-c) => ((-a) "/\" -b)) = Top L1 by A65,A2; then
     -c < -(a "\/" b) by A2;
     hence thesis by A65,A64,A17,A6;
   end;
A66: for a, b, c being Element of L1 st
         c <= a & c <= b holds c <= a "/\" b
   proof
     let a, b, c be Element of L1;
     assume
A67: c <= a & c <= b; then
A68: c < a & -a < -c by A6;
A69: c < b & -b < -c by A67,A6;
     ((-a) => (-c)) => (((-b) => (-c)) => (((-a) "\/" -b) => -c)) = Top L1
       by A2; then
     (((-b) => (-c)) => (((-a) "\/" -b) => -c)) = Top L1 by A68,A2; then
     ((((-a) "\/" -b) => -c)) = Top L1 by A69,A2; then
     -(a "/\" b) < -c by A2;
     hence thesis by A69,A68,A22,A6;
   end;

A70: for a,b being Element of L1 holds
      b "\/" a <= a "\/" b
   proof
     let a,b be Element of L1;
A71: a <= a "\/" b by A28;
     b <= a "\/" b by A31;
     hence thesis by A71,A62;
   end;
A72: for a,b being Element of L1 holds
      a "\/" b = b "\/" a
   proof
     let a,b be Element of L1;
A73:  a "\/" b <= b "\/" a by A70;
     b "\/" a <= a "\/" b by A70;
     hence thesis by A73,A40;
   end;
A74: for a,b being Element of L1 holds
      a "/\" b <= b "/\" a
   proof
     let a,b be Element of L1;
A75: a "/\" b <= a by A25;
     a "/\" b <= b by A34;
     hence thesis by A75,A66;
   end;
   for a,b being Element of L1 holds
     a "/\" b = b "/\" a
   proof
     let a,b be Element of L1;
A76: a "/\" b <= b "/\" a by A74;
     b "/\" a <= a "/\" b by A74;
     hence thesis by A40,A76;
   end; then
   reconsider L1 as DeMorgan meet-commutative join-commutative
     non empty NelsonStr by A72,LATTICES:def 4, def 6;

A77: for a, b, c being Element of L1 holds
     a <= b implies a "\/" c <= b "\/" c
   proof
      let a,b,c be Element of L1;
      assume a <= b; then
A78:  a < b & -b < -a by A6; then
      (-b) "/\" -c < (-a) "/\" -c by A57; then
      - (b "\/" c) < (-a) "/\" -c by A2; then
      - (b "\/" c) < - (a "\/" c) by A2;
      hence thesis by A78,A58,A6;
    end;
    set d = -Top L1;
A79: for a being Element of L1 holds d "/\" a = d & a "/\" d = d by A12;
    for a, b being Element of L1 holds
      b = (a "/\" b) "\/" b
    proof
      let a, b be Element of L1;
A80:  b <= (a "/\" b) "\/" b by A31;
      a "/\" b <= b & b <= b by A40,A25; then
      (a "/\" b) "\/" b <= b by A62;
      hence thesis by A80,A40;
    end; then
A81: L1 is meet-absorbing;
    for a,b being Element of L1 holds
      a "/\" (a "\/" b) = a
    proof
      let a, b be Element of L1;
      a <= a & a <= a "\/" b by A40,A31;
      hence thesis;
    end; then
A82: L1 is join-absorbing;
A83: for a, b, c being Element of L1 holds
    b <= c implies a "/\" b <= a "/\" c
    proof
      let a, b, c be Element of L1;
      assume
A84:  b <= c; then
A85:  a "/\" b < a "/\" c by A57,A6;
      -(a "/\" c) < -(a "/\" b)
      proof
        -c < -b by A84,A6; then
        (-a) "\/" (-c) < (-a) "\/" (-b) by A58; then
        -(a "/\" c) < (-a) "\/" (-b) by A2;
        hence thesis by A2;
      end;
      hence thesis by A85,A6;
    end;
A86: for a,b,c being Element of L1 st a <= b & b <= c holds a <= c
    proof
      let a,b,c be Element of L1;
      assume a <= b & b <= c; then
      a < b & b < c & -c < -b & -b < -a by A6; then
      a < c & (-c) < -a by A14;
      hence thesis by A6;
    end;
A87: for a,b,c being Element of L1 holds
      a "/\" (b "/\" c) = (a "/\" b) "/\" c
    proof
      let a,b,c be Element of L1;
A88:  a "/\" (b "/\" c) <= a "/\" b by A34,A83;
A89:  a "/\" (b "/\" c) <= b "/\" c by A34;
      b "/\" c <= c by A34; then
      a "/\" (b "/\" c) <= c by A86,A89; then
A90:  a "/\" (b "/\" c) <= (a "/\" b) "/\" c by A88,A66;
A91:  a "/\" b <= a by A25;
      (a "/\" b) "/\" c <= a "/\" b by A25; then
A92:  (a "/\" b) "/\" c <= a by A91,A86;
      (a "/\" b) "/\" c <= b "/\" c by A25,A83; then
      (a "/\" b) "/\" c <= a "/\" (b "/\" c) by A92,A66;
      hence thesis by A90,A40;
    end;
    for a,b,c being Element of L1 holds
      a "\/" (b "\/" c) = (a "\/" b) "\/" c
    proof
      let a,b,c be Element of L1;
A93:  a <= a "\/" b by A28;
      a "\/" b <= (a "\/" b) "\/" c by A28; then
A94:  a <= (a "\/" b) "\/" c by A86,A93;
      b "\/" c <= (a "\/" b) "\/" c by A28,A77; then
A95:  a "\/" (b "\/" c) <= (a "\/" b) "\/" c by A94,A62;
A96:  c <= b "\/" c by A28;
      b "\/" c <= a "\/" (b "\/" c) by A28; then
A97:  c <= a "\/" (b "\/" c) by A96,A86;
      a "\/" b <= a "\/" (b "\/" c) by A28,A77; then
      (a "\/" b) "\/" c <= a "\/" (b "\/" c) by A97,A62;
      hence thesis by A95,A40;
    end; then
    L1 is join-associative meet-associative by A87; then
    reconsider L1 as Lattice-like
      lower-bounded DeMorgan non empty NelsonStr
        by A81,A79,A82,LATTICES:def 13;
    set c = Top L1;
    for a being Element of L1 holds c "\/" a = c & a "\/" c = c
    proof
      let a be Element of L1;
      a <= c by A8;
      hence thesis by LATTICES:4,def 3;
    end; then
    L is upper-bounded; then
    reconsider L1 as DeMorgan involutive bounded Lattice-like
        non empty NelsonStr by A2;
A98: L1 is distributive
    proof
      let a, b, c be Element of L1;
A99:  b < a => ((a "/\" b) "\/" (a "/\" c)) by A52,A24;
      c < a => ((a "/\" b) "\/" (a "/\" c)) by A52,A24; then
A100: b "\/" c < a => ((a "/\" b) "\/" (a "/\" c)) by A99,A17;
A101: for a, b, c being Element of L1 holds
        a "/\" (b "\/" c) < (a "/\" b) "\/" (a "/\" c)
      proof
        let a, b, c be Element of L1;
A102:   b < a => ((a "/\" b) "\/" (a "/\" c)) by A52,A24;
        c < a => ((a "/\" b) "\/" (a "/\" c)) by A52,A24; then
        b "\/" c < a => ((a "/\" b) "\/" (a "/\" c)) by A102,A17;
        hence thesis by A52;
      end;
A103: ((-a) "\/" (-b)) "/\" ((-a) "\/" (-c)) <
    (((-a) "\/" (-b)) "/\" (-a)) "\/" (((-a) "\/" (-b)) "/\" (-c)) by A101;
A104: (((-a) "\/" (-b)) "/\" (-a)) "\/" (((-a) "\/" (-b)) "/\" (-c)) =
    (-a) "\/" (((-a) "\/" (-b)) "/\" (-c)) by LATTICES:def 9;
    (-a) "\/"((-c) "/\" ((-a) "\/" (-b))) <
       (-a) "\/" (((-c) "/\" (-a)) "\/" ((-c) "/\" (-b))) by A101,A58; then
A105:((-a) "\/" (-b)) "/\" ((-a) "\/" (-c)) <
        (-a) "\/" (((-a) "/\" (-c)) "\/" ((-b) "/\" (-c))) by A16,A103,A104;
    (-a) "\/" (((-a) "/\" (-c)) "\/" ((-b) "/\" (-c))) =
      ((-a) "\/" ((-a) "/\" (-c))) "\/" ((-b) "/\" (-c)) by LATTICES:def 5
      .= (-a) "\/" ((-b) "/\" (-c)) by LATTICES:def 8
      .= (-a) "\/" -(b "\/" c) by A2; then
    ((-a) "\/" (-b)) "/\" (-(a "/\" c)) < (-a) "\/" (-(b "\/" c))
        by A2,A105; then
    (-(a "/\" b)) "/\" (-(a "/\" c)) < (-a) "\/" (-(b "\/" c))
        by A2; then
    (-(a "/\" b)) "/\" (-(a "/\" c)) < -(a "/\" (b "\/" c)) by A2; then
    -((a "/\" b) "\/" (a "/\" c)) < -(a "/\" (b "\/" c)) by A2; then
A106: a "/\" (b "\/" c) <= (a "/\" b) "\/" (a "/\" c) by A6,A100,A52;
A107: a "/\" b <= a "/\" (b "\/" c) by A28,A83;
    a "/\" c <= a "/\" (b "\/" c) by A28,A83; then
    (a "/\" b) "\/" (a "/\" c) <= a "/\" (b "\/" c) by A62,A107;
    hence thesis by A106,A40;
  end;
  reconsider L1 as DeMorgan involutive bounded distributive Lattice-like
    non empty NelsonStr by A98;
A108:  L1 is satisfying_N5
  proof
    let a, b be Element of L1;
A109: a =-> b = Top L1 implies a "/\" b = a
    proof
      assume a =-> b = Top L1; then
      a <= b by A1;
      hence thesis;
    end;
    a "/\" b = a implies a =-> b = Top L1
    proof
      assume a "/\" b = a; then
      a <= b;
      hence thesis by A1;
    end;
    hence thesis by A109;
  end;
A110: L1 is satisfying_N8
  proof
    let a, b be Element of L1;
    ((-(a => b)) => (a "/\" -b)) "/\" ((a "/\" -b) =>
    (-(a => b))) = Top L1 by A2;
    hence thesis by A3;
  end;
A111: L1 is satisfying_N9
  proof
    let a, b be Element of L1;
    ((-(a => b)) => (a "/\" -b)) "/\" ((a "/\" -b) =>
       (-(a => b))) = Top L1 by A2;
    hence thesis by A3;
  end;
A112: L1 is satisfying_N10
  proof
    let a be Element of L1;
    ((-!a) => a) "/\" (a => (-!a)) = Top L1 by A2;
    hence thesis by A3;
  end;
A113: L1 is satisfying_N11
  proof
    let a be Element of L1;
    ((-!a) => a) "/\" (a => (-!a)) = Top L1 by A2;
    hence thesis by A3;
  end;
A114: L1 is satisfying_N12
  proof
    let a, b be Element of L1;
    (-a) < a => b by A2;
    hence thesis by A52;
  end;

A115: for a, b, c being Element of L1 holds !(Top L1) = -(Top L1)
  proof
    let a, b, c be Element of L1;
A116: -(Top L1) <= !(Top L1) by A10;
    !(Top L1) <= -(Top L1)
    proof
      (!(a => a)) => -( Top L1) = (Top L1) by A2; then
A117:   !(Top L1) < -(Top L1) by A19;
A118:   -(-(Top L1)) = (Top L1) by A2;
      (Top L1) < (-(!(Top L1))) by A112;
      hence thesis by A117,A118,A6;
    end;
    hence thesis by A116,A40;
  end;

A119: for a, b being Element of L1 holds a => !b = b => !a
    proof
      let a, b be Element of L1;
A120: a => !b < b => !a by A2;
A121: -(b => !a) < b "/\" -(!a) by A111;
      b "/\" -(!a) < b "/\" a by A113,A57; then
A122: -(b => !a) < a "/\" b by A121,A16;
A123: a "/\" b < a "/\" -(!b) by A112,A57;
      a "/\" -(!b) < -(a => !b) by A110; then
      a "/\" b < -(a => !b) by A123,A16; then
      -(b => !a) < -(a => !b) by A16,A122; then
A124: a => !b <= b => !a by A120,A6;
A125: b => !a < a => !b by A2;
A126: -(a => !b) < a "/\" -(!b) by A111;
      a "/\" -(!b) < a "/\" b by A113,A57; then
A127: -(a => !b) < b "/\" a by A126,A16;
A128: b "/\" a < b "/\" -(!a) by A112,A57;
      b "/\" -(!a) < -(b => !a) by A110; then
      b "/\" a < -(b => !a) by A128,A16; then
      -(a => !b) < -(b => !a) by A16,A127; then
      b => !a <= a => !b by A125,A6;
      hence thesis by A124,A40;
    end;
    L1 is satisfying_N13
    proof
      let a be Element of L1;
      (!a) = (Top L1) => (!a) by A2
          .= a => !(Top L1) by A119
          .= a => -(Top L1) by A115;
      hence thesis;
    end;
    hence thesis by A108,A42,A52,A2,A17,A22,A110,A111,A112,A113,A114,A16;
  end;
