reserve a,b,c,k,m,n for Nat;
reserve i,j,x,y for Integer;
reserve p,q for Prime;
reserve r,s for Real;

theorem
  { n where n is 3-gonal Nat: n is 5-gonal } is infinite
  proof
    set A = { n where n is 3-gonal Nat: n is 5-gonal };
    set B = { [x,y] where x,y is positive Nat: y*(3*y-1) - x*(x+1) = 0 };
    defpred P[object,object] means
    ex n being positive Nat st n = $1`1 & $2 = Polygon(3,n);
A1: for e being object st e in B ex u being object st P[e,u]
    proof
      let e be object;
      assume e in B;
      then consider x,y being positive Nat such that
A2:   e = [x,y] and y*(3*y-1) - x*(x+1) = 0;
      take u = Polygon(3,x), n = x;
      thus n = e`1 by A2;
      thus u = Polygon(3,n);
    end;
    consider f being Function such that
A3: dom f = B and
A4: for e being object st e in B holds P[e,f.e] from CLASSES1:sch 1(A1);
A5: f is one-to-one
    proof
      let x1,x2 be object such that
A6:   x1 in dom f and
A7:   x2 in dom f and
A8:   f.x1 = f.x2;
      consider n1 being positive Nat such that
A9:  n1 = x1`1 and
A10:  f.x1 = Polygon(3,n1) by A3,A4,A6;
      consider n2 being positive Nat such that
A11:  n2 = x2`1 & f.x2 = Polygon(3,n2) by A3,A4,A7;
      consider c1,d1 being positive Nat such that
A12:  x1 = [c1,d1] and
A13:  d1*(3*d1-1) - c1*(c1+1) = 0 by A3,A6;
      consider c2,d2 being positive Nat such that
A14:  x2 = [c2,d2] and
A15:  d2*(3*d2-1) - c2*(c2+1) = 0 by A3,A7;
A16:  3*(d1*d1-d2*d2) = d1-d2 by A8,A9,A10,A11,A12,A13,A14,A15;
      now
        assume d1 <> d2;
        then per cases by XXREAL_0:1;
        suppose d1 < d2;
          then consider z being positive Nat such that
A17:      d2 = d1+z by Th1;
          3*(d1*d1-(d1+z)*(d1+z)) = d1-d1-z by A16,A17;
          then 3*(2*d1+z)*z = 1*z;
          then 3*(2*d1+z) = 1 by XCMPLX_1:5;
          hence contradiction by Lm20;
        end;
        suppose d2 < d1;
          then consider z being positive Nat such that
A18:      d1 = d2+z by Th1;
          3*(2*d2+z)*z = 1*z by A16,A18;
          then 3*(2*d2+z) = 1 by XCMPLX_1:5;
          hence contradiction by Lm20;
        end;
      end;
      hence thesis by A8,A9,A10,A11,A12,A14,Th67;
    end;
    rng f c= A
    proof
      let t be object;
      assume t in rng f;
      then consider s being object such that
A19:  s in dom f and
A20:  f.s = t by FUNCT_1:def 3;
      consider x,y being positive Nat such that
A21:  s = [x,y] & y*(3*y-1) - x*(x+1) = 0 by A3,A19;
A22:  P[s,f.s] by A3,A4,A19;
      then f.s = Polygon(5,y) by A21;
      hence thesis by A20,A22;
    end;
    hence thesis by A3,A5,Th68,CARD_1:59;
  end;
