reserve SAS for AffinPlane;

theorem Th1:
  SAS is Pappian implies for a1,a2,a3,b1,b2,b3 being Element of SAS
  holds ( a1,a2 // a1,a3 & b1,b2 // b1,b3 & a1,b2 // a2,b1 & a2,b3 // a3,b2
  implies a3,b1 // a1,b3 )
proof
  assume
A1: SAS is Pappian;
  let a1,a2,a3,b1,b2,b3 be Element of SAS such that
A2: a1,a2 // a1,a3 and
A3: b1,b2 // b1,b3 and
A4: a1,b2 // a2,b1 and
A5: a2,b3 // a3,b2;
  LIN a1,a2,a3 by A2,AFF_1:def 1;
  then consider M being Subset of SAS such that
A6: M is being_line and
A7: a1 in M and
A8: a2 in M and
A9: a3 in M by AFF_1:21;
  LIN b1,b2,b3 by A3,AFF_1:def 1;
  then consider N being Subset of SAS such that
A10: N is being_line and
A11: b1 in N and
A12: b2 in N and
A13: b3 in N by AFF_1:21;
A14: now
    assume that
A15: M<>N and
A16: not M // N;
    consider o being Element of SAS such that
A17: o in M and
A18: o in N by A6,A10,A16,AFF_1:58;
A19: now
      assume that
A20:  o<>a1 and
A21:  b1<>o and
A22:  o=b3;
A23:  now
        assume a2<>o;
        then b2 in M by A5,A6,A8,A9,A17,A22,AFF_1:48;
        then b2=o by A6,A10,A12,A15,A17,A18,AFF_1:18;
        then b1 in M by A4,A6,A7,A8,A17,A20,AFF_1:48;
        hence thesis by A6,A7,A9,A17,A22,AFF_1:51;
      end;
      now
        assume a2=o;
        then o,b1 // b2,a1 by A4,AFF_1:4;
        then a1 in N by A10,A11,A12,A13,A21,A22,AFF_1:48;
        hence thesis by A6,A7,A10,A15,A17,A18,A20,AFF_1:18;
      end;
      hence thesis by A23;
    end;
A24: now
      assume that
      o<>a1 and
A25:  b1=o;
A26:  o,a2 // a1,b2 by A4,A25,AFF_1:4;
A27:  now
        assume a2<>o;
        then b2 in M by A6,A7,A8,A17,A26,AFF_1:48;
        then b2=o by A6,A10,A12,A15,A17,A18,AFF_1:18;
        then
A28:    o,a3 // a2,b3 by A5,AFF_1:4;
        now
          assume o<>a3;
          then b3 in M by A6,A8,A9,A17,A28,AFF_1:48;
          hence thesis by A6,A7,A9,A17,A25,AFF_1:51;
        end;
        hence thesis by A25,AFF_1:3;
      end;
      now
        assume
A29:    a2=o;
        now
          assume
A30:      b3<>o;
          o,b3 // b2,a3 by A5,A29,AFF_1:4;
          then a3 in N by A10,A12,A13,A18,A30,AFF_1:48;
          then b1=a3 by A6,A9,A10,A15,A17,A18,A25,AFF_1:18;
          hence thesis by AFF_1:3;
        end;
        hence thesis by A6,A7,A9,A17,A25,AFF_1:51;
      end;
      hence thesis by A27;
    end;
A31: now
      assume
A32:  o=a1;
      then
A33:  o,b2 // b1,a2 by A4,AFF_1:4;
A34:  now
        assume b2<>o;
        then a2 in N by A10,A11,A12,A18,A33,AFF_1:48;
        then a2=o by A6,A8,A10,A15,A17,A18,AFF_1:18;
        then
A35:    o,b3 // b2,a3 by A5,AFF_1:4;
        now
          assume o<>b3;
          then a3 in N by A10,A12,A13,A18,A35,AFF_1:48;
          hence thesis by A10,A11,A13,A18,A32,AFF_1:51;
        end;
        hence thesis by A32,AFF_1:3;
      end;
      now
        assume
A36:    b2=o;
        now
          assume
A37:      a3<>o;
          o,a3 // a2,b3 by A5,A36,AFF_1:4;
          then b3 in M by A6,A8,A9,A17,A37,AFF_1:48;
          then a1=b3 by A6,A10,A13,A15,A17,A18,A32,AFF_1:18;
          hence thesis by AFF_1:3;
        end;
        hence thesis by A10,A11,A13,A18,A32,AFF_1:51;
      end;
      hence thesis by A34;
    end;
A38: now
      assume that
A39:  o<>a1 and
A40:  b1<>o and
      o<>b3 and
A41:  o=a3;
A42:  o,b2 // b3,a2 by A5,A41,AFF_1:4;
A43:  now
        assume b2<>o;
        then a2 in N by A10,A12,A13,A18,A42,AFF_1:48;
        then a2=o by A6,A8,A10,A15,A17,A18,AFF_1:18;
        then o,b1 // b2,a1 by A4,AFF_1:4;
        then a1 in N by A10,A11,A12,A18,A40,AFF_1:48;
        hence thesis by A10,A11,A13,A18,A41,AFF_1:51;
      end;
      now
        assume b2=o;
        then b1 in M by A4,A6,A7,A8,A9,A39,A41,AFF_1:48;
        hence thesis by A6,A10,A11,A15,A17,A18,A40,AFF_1:18;
      end;
      hence thesis by A43;
    end;
    now
      assume that
A44:  o<>a1 & o<>b1 and
A45:  o<>b3 and
A46:  o<>a3;
A47:  o<>b2
      proof
        assume o=b2;
        then o,a3 // a2,b3 by A5,AFF_1:4;
        then b3 in M by A6,A8,A9,A17,A46,AFF_1:48;
        hence contradiction by A6,A10,A13,A15,A17,A18,A45,AFF_1:18;
      end;
      o<>a2
      proof
        assume o=a2;
        then o,b3 // b2,a3 by A5,AFF_1:4;
        then a3 in N by A10,A12,A13,A18,A45,AFF_1:48;
        hence contradiction by A6,A9,A10,A15,A17,A18,A46,AFF_1:18;
      end;
      then a1,b3 // a3,b1 by A1,A4,A5,A6,A7,A8,A9,A10,A11,A12,A13,A15,A17,A18
,A44,A45,A46,A47,AFF_2:def 2;
      hence thesis by AFF_1:4;
    end;
    hence thesis by A31,A24,A19,A38;
  end;
A48: SAS is satisfying_pap by A1,AFF_2:9;
  now
    assume M<>N & M // N;
    then a1,b3 // a3,b1 by A48,A4,A5,A6,A7,A8,A9,A10,A11,A12,A13,AFF_2:def 13;
    hence thesis by AFF_1:4;
  end;
  hence thesis by A6,A7,A9,A11,A13,A14,AFF_1:51;
end;
