reserve AP for AffinPlane,
  a,a9,b,b9,c,c9,x,y,o,p,q,r,s for Element of AP,
  A,C,C9,D,K,M,N,P,T for Subset of AP;

theorem
  AP is Pappian iff AP is satisfying_PAP_1
proof
    hereby assume
A1: AP is Pappian;
    thus AP is satisfying_PAP_1
    proof
      let M,N,o,a,b,c,a9,b9,c9;
      assume that
A2:  M is being_line and
A3:  N is being_line and
A4:  M<>N and
A5:  o in M and
A6:  o in N and
A7:  o<>a and
      o<>a9 and
A8:  o<>b and
A9:  o<>b9 and
A10:  o<>c and
A11:  o<>c9 and
A12:  a in M and
A13:  b in M and
A14:  c in M and
A15:  b9 in N and
A16:  c9 in N and
A17:  a,b9 // b,a9 and
A18:  b,c9 // c,b9 and
A19:  a,c9 // c,a9 and
A20:  b<>c;
A21:  a<>c9 by A2,A3,A4,A5,A6,A7,A12,A16,AFF_1:18;
A22:  b<>a9
      proof
        assume b=a9;
        then c,b // a,c9 by A19,AFF_1:4;
        then c9 in M by A2,A12,A13,A14,A20,AFF_1:48;
        hence contradiction by A2,A3,A4,A5,A6,A11,A16,AFF_1:18;
      end;
      not b,a9 // N
      proof
        assume
A23:    b,a9 // N;
        b,a9 // a,b9 by A17,AFF_1:4;
        then a,b9 // N by A22,A23,AFF_1:32;
        then b9,a // N by AFF_1:34;
        then a in N by A3,A15,AFF_1:23;
        hence contradiction by A2,A3,A4,A5,A6,A7,A12,AFF_1:18;
      end;
      then consider x such that
A24:  x in N and
A25:  LIN b,a9,x by A3,AFF_1:59;
A26:  b,a9 // b,x by A25,AFF_1:def 1;
A27:  o<>x
      proof
        assume o=x;
        then b,o // a,b9 by A17,A22,A26,AFF_1:5;
        then b9 in M by A2,A5,A8,A12,A13,AFF_1:48;
        hence contradiction by A2,A3,A4,A5,A6,A9,A15,AFF_1:18;
      end;
      a,b9 // b,x by A17,A22,A26,AFF_1:5;
      then a,c9 // c,x by A1,A2,A3,A4,A5,A6,A7,A8,A9,A10,A11,A12,A13,A14,A15
,A16,A18,A24,A27;
      then c,a9 // c,x by A19,A21,AFF_1:5;
      then LIN c,a9,x by AFF_1:def 1;
      then
A28:  LIN a9,x,c by AFF_1:6;
A29:  LIN a9,x,x by AFF_1:7;
      assume
A30:  not a9 in N;
      LIN a9,x,b by A25,AFF_1:6;
      then x in M by A2,A13,A14,A20,A30,A24,A28,A29,AFF_1:8,25;
      hence contradiction by A2,A3,A4,A5,A6,A24,A27,AFF_1:18;
    end;
  end;
    assume
A31: AP is satisfying_PAP_1;
      let M,N,o,a,b,c,a9,b9,c9;
      assume that
A32:   M is being_line and
A33:   N is being_line and
A34:   M<>N & o in M & o in N and
A35:   o<>a and
A36:   o<>a9 and
A37:   o<>b and
A38:   o<>b9 and
A39:  o<>c & o<>c9 and
A40:  a in M and
A41:  b in M and
A42:  c in M and
A43:  a9 in N and
A44:  b9 in N and
A45:  c9 in N and
A46:  a,b9 // b,a9 and
A47:  b,c9 // c,b9;
      set A=Line(a,c9), P=Line(b,a9);
A48:  b<>a9 by A32,A33,A34,A36,A41,A43,AFF_1:18;
      then
A49:  b in P by AFF_1:24;
      assume
A50:  not a,c9 // c,a9;
      then
A51:  a<>c9 by AFF_1:3;
      then
A52:  a in A & c9 in A by AFF_1:24;
A53:  a9 in P by A48,AFF_1:24;
      A is being_line by A51,AFF_1:24;
      then consider K such that
A54:  c in K and
A55:  A // K by AFF_1:49;
A56:  b<>c
      proof
        assume
A57:    b=c;
        then LIN b,c9,b9 by A47,AFF_1:def 1;
        then LIN b9,c9,b by AFF_1:6;
        then b9=c9 or b in N by A33,A44,A45,AFF_1:25;
        hence contradiction by A32,A33,A34,A37,A41,A46,A50,A57,AFF_1:18;
      end;
A58:  b9<>c9
      proof
        assume b9=c9;
        then b9,b // b9,c by A47,AFF_1:4;
        then LIN b9,b,c by AFF_1:def 1;
        then LIN b,c,b9 by AFF_1:6;
        then b9 in M by A32,A41,A42,A56,AFF_1:25;
        hence contradiction by A32,A33,A34,A38,A44,AFF_1:18;
      end;
A59:  not P // K
      proof
        assume P // K;
        then P // A by A55,AFF_1:44;
        then b,a9 // a,c9 by A52,A49,A53,AFF_1:39;
        then a,b9 // a,c9 by A46,A48,AFF_1:5;
        then LIN a,b9,c9 by AFF_1:def 1;
        then LIN b9,c9,a by AFF_1:6;
        then a in N by A33,A44,A45,A58,AFF_1:25;
        hence contradiction by A32,A33,A34,A35,A40,AFF_1:18;
      end;
A60:  P is being_line by A48,AFF_1:24;
      K is being_line by A55,AFF_1:36;
      then consider x such that
A61:  x in P and
A62:  x in K by A60,A59,AFF_1:58;
A63:  a,c9 // c,x by A52,A54,A55,A62,AFF_1:39;
      LIN b,x,a9 by A60,A49,A53,A61,AFF_1:21;
      then b,x // b,a9 by AFF_1:def 1;
      then a,b9 // b,x by A46,A48,AFF_1:5;
      then x in N by A31,A32,A33,A34,A35,A37,A38,A39,A40,A41,A42,A44,A45,A47
,A56,A63;
      then N=P by A33,A43,A50,A60,A53,A61,A63,AFF_1:18;
      hence contradiction by A32,A33,A34,A37,A41,A49,AFF_1:18;
end;
