
theorem Th6:
  for n be Nat for p,q be Element of n-tuples_on NAT
holds p <> q implies ex i be Element of NAT st i in Seg n & p.i <> q.i &
   for k be Nat st 1 <= k & k < i holds p.k = q.k
proof
  defpred P[Nat] means for p,q be Element of $1-tuples_on NAT holds p <> q
  implies ex i be Element of NAT st i in Seg $1 & p.i <> q.i &
   for k be Nat st 1 <= k & k < i holds p.k = q.k;
A1: for n be Nat st P[n] holds P[n+1]
  proof
    let n be Nat;
    assume
A2: for p,q be Element of n-tuples_on NAT holds p <> q implies ex i be
Element of NAT st i in Seg n & p.i <> q.i & for k be Nat st 1 <= k &
    k < i holds p.k = q.k;
    let p,q be Element of (n+1)-tuples_on NAT;
    consider t1 be Element of n-tuples_on NAT, d1 be Element of NAT such that
A3: p = t1^<*d1*> by FINSEQ_2:117;
    assume
A4: p <> q;
    consider t2 be Element of n-tuples_on NAT, d2 be Element of NAT such that
A5: q = t2^<*d2*> by FINSEQ_2:117;
A6: len t1 = n by CARD_1:def 7;
A7: len t2 = n by CARD_1:def 7;
    per cases;
    suppose
      t1 <> t2;
      then consider i be Element of NAT such that
A8:   i in Seg n and
A9:   t1.i <> t2.i and
A10:  for k be Nat st 1 <= k & k < i holds t1.k = t2.k by A2;
      take i;
      thus i in Seg (n+1) by A8,FINSEQ_2:8;
      i in dom t1 by A6,A8,FINSEQ_1:def 3;
      then
A11:  p.i = t1.i by A3,FINSEQ_1:def 7;
      i in dom t2 by A7,A8,FINSEQ_1:def 3;
      hence p.i <> q.i by A5,A9,A11,FINSEQ_1:def 7;
      let k be Nat;
      assume that
A12:  1 <= k and
A13:  k < i;
      i <= n by A8,FINSEQ_1:1;
      then
A14:  k <= n by A13,XXREAL_0:2;
      then
A15:  k in dom t1 by A6,A12,FINSEQ_3:25;
A16:  k in dom t2 by A7,A12,A14,FINSEQ_3:25;
      t1.k = t2.k by A10,A12,A13;
      hence p.k = t2.k by A3,A15,FINSEQ_1:def 7
        .= q.k by A5,A16,FINSEQ_1:def 7;
    end;
    suppose
A17:  t1 = t2;
      take i = n+1;
      thus i in Seg (n+1) by FINSEQ_1:4;
      p.i = d1 by A3,FINSEQ_2:116;
      hence p.i <> q.i by A3,A5,A4,A17,FINSEQ_2:116;
      let k be Nat;
      assume that
A18:  1 <= k and
A19:  k < i;
A20:  k <= n by A19,NAT_1:13;
      then
A21:  k in dom t2 by A7,A18,FINSEQ_3:25;
      k in dom t1 by A6,A18,A20,FINSEQ_3:25;
      hence p.k = t2.k by A3,A17,FINSEQ_1:def 7
        .= q.k by A5,A21,FINSEQ_1:def 7;
    end;
  end;
A22: P[0];
  thus for n be Nat holds P[n] from NAT_1:sch 2(A22,A1);
end;
