 reserve V for Z_Module;
 reserve W for Subspace of V;
 reserve v, u for Vector of V;
 reserve i for Element of INT.Ring;

theorem LmISRank42:
  for V being torsion-free Z_Module,
  W1, W2 being finite-rank free Subspace of V, I being Basis of W1
  st (for v being Vector of V st v in I holds (W1 /\ W2) /\ Lin{v} <> (0).V)
  holds rank(W1 + W2) = rank(W2)
  proof
    let V be torsion-free Z_Module;
    defpred P[Nat] means
    for W1, W2 being finite-rank free Subspace of V, I being Basis of W1
    st (for v being Vector of V st v in I holds (W1 /\ W2) /\ Lin{v} <> (0).V)
    & rank(W1) = $1 holds rank(W1 + W2) = rank(W2);
    A1: P[0]
    proof
      let W1, W2 be finite-rank free Subspace of V,
      I be Basis of W1 such that
      B1: (for v being Vector of V st v in I holds
      (W1 /\ W2) /\ Lin{v} <> (0).V) & rank(W1) = 0;
      B2: (Omega).W1 = (0).W1 by B1,ZMODUL05:1
      .= (0).V by ZMODUL01:51;
      reconsider W1s = (Omega).W1, W2s = (Omega).W2
      as strict finite-rank free Subspace of V by ZMODUL01:42;
      thus rank(W1 + W2) = rank(W1s + W2s) by ZMODUL04:22
      .= rank(W2s) by ZMODUL01:99,B2
      .= rank(W2) by ZMODUL05:4;
    end;
    A2: for n being Nat st P[n] holds P[n+1]
    proof
      let n be Nat such that
      B1: P[n];
      let W1, W2 be finite-rank free Subspace of V,
      I be Basis of W1 such that
      B2: (for v being Vector of V st v in I holds
      (W1 /\ W2) /\ Lin{v} <> (0).V) & rank(W1) = n+1;
      card(I) > 0 by ZMODUL03:def 5,B2;
      then I <> {}(the carrier of W1);
      then consider u be object such that
      B3: u in I by XBOOLE_0:7;
      reconsider u as Vector of W1 by B3;
      reconsider uu = u as Vector of V by ZMODUL01:25;
      B4: I is linearly-independent by VECTSP_7:def 3;
      reconsider II = I as linearly-independent Subset of V
      by ZMODUL03:15,VECTSP_7:def 3;
      I \ {u} is linearly-independent by B4,XBOOLE_1:36,ZMODUL02:56;
      then reconsider Iu = I \ {u} as linearly-independent Subset of V
      by ZMODUL03:15;
      BX2X: (Omega).W1 = Lin(I) by VECTSP_7:def 3
      .= Lin(II) by ZMODUL03:20;
      then BX2: (Omega).W1 = Lin(Iu) + Lin{uu} & Lin(Iu) /\ Lin{uu} = (0).V &
      Lin(Iu) is free & Lin{uu} is free & uu <> 0.V by B3,ThLin8;
      reconsider LIu = Lin(Iu) as finite-rank free Subspace of V;
      B5: Iu is Basis of Lin(Iu) by ThLin7;
      card(Iu) = card(I) - card{u} by B3,ZFMISC_1:31,CARD_2:44
      .= rank(W1) - card{u} by ZMODUL03:def 5
      .= n + 1 - 1 by B2,CARD_1:30
      .= n;
      then B6: rank(LIu) = n by B5,ZMODUL03:def 5;
      B7X: for v being Vector of V st v in Iu holds
      (Lin(Iu) /\ W2) /\ Lin{v} <> (0).V
      proof
        let v be Vector of V such that
        C1: v in Iu;
        v in I by C1,TARSKI:def 3,XBOOLE_1:36;
        then (W1 /\ W2) /\ Lin{v} <> (0).V by B2;
        then W1 /\ (W2 /\ Lin{v}) <> (0).V by ZMODUL01:104;
        then C2: W2 /\ Lin{v} <> (0).V by ZMODUL01:107;
        C3: v <> 0.V by C1,ZMODUL02:57;
        C4: v in Lin{v} by ZMODUL02:65,ZFMISC_1:31;
        v in LIu by C1,ZMODUL02:65;
        then LIu /\ Lin{v} <> (0).V by C3,ZMODUL02:66,C4,ZMODUL01:94;
        hence thesis by C2,LmISRank21;
      end;
      (W1 /\ W2) /\ Lin{uu} = W1 /\ (W2 /\ Lin{uu}) by ZMODUL01:104;
      then W2 /\ Lin{uu} <> (0).V by ZMODUL01:107,B2,B3;
      then (LIu + W2) /\ Lin{uu} <> (0).V by ThIS1;
      then B8: rank((LIu + W2) + Lin{uu}) = rank(LIu + W2) by BX2,LmRank2
      .= rank(W2) by B7X,B1,B5,B6;
      reconsider W1s = (Omega).W1, W2s = (Omega).W2
      as strict finite-rank free Subspace of V by ZMODUL01:42;
      B9: (Omega).W1s = W1s;
      (LIu + W2) + Lin{uu} = (Lin{uu} + LIu) + W2 by ZMODUL01:96
      .= W1s + W2 by BX2X,B3,ThLin8
      .= W1s + W2s by B9,ZMODUL04:22
      .= W1 + W2 by ZMODUL04:22;
      hence thesis by B8;
    end;
    A3: for n being Nat holds P[n] from NAT_1:sch 2(A1,A2);
    let W1, W2 be finite-rank free Subspace of V,
    I be Basis of W1 such that
    A4: for v being Vector of V st v in I holds (W1 /\ W2) /\ Lin{v} <> (0).V;
    set rk = rank(W1);
    P[rk] by A3;
    hence thesis by A4;
  end;
