Proof of the Inequality Shown in Equation (5.2)

Now substituting equations (5.26), (5.46), (5.62) and (5.76) in equation (5.96), we have:

dim(A)≤(q−1)(dim(W) +dim(X) +dim(Y) + 2dim(Z)) +

q−1

X

i=1

dim(Vi)−2(q−1)dim(A)

−2(q−1)dim(C)−

q−1

X

i=1

2dim(B_i) + (7q−6)codim_W(W⁰) + (6q−5)codim_X(X⁰)

+

q−1

X

i=1

(2q)codimVi(V_i⁰) + (3q−3)codimY(Y⁰) + (4q−3)codimZ(Z⁰) + (2q−1)codimA(A⁰) + (q−1)codim_C(C⁰) +

q−1

X

i=1

2codim_B(B_i⁰) +

q−1

X

i=1

codim_C(C_i⁰) + (5q−4)(dim(A)−dim(A, B₁, . . . , Bq−1, C))

+ (6q−5)(

q−1

X

i=1

dim(B_i) +dim(C))−(q−1)dim(B₁, . . . , Bq−1, C).

Substituting values from equations (5.11), (5.12), (5.10), (5.13), (5.14), (5.6), (5.7), (5.8), and (5.9), we get:

dim(A)≤(q−1)(dim(W) +dim(X) +dim(Y) + 2dim(Z)) +

q−1

X

i=1

dim(V_i)−2(q−1)dim(A)

−2(q−1)dim(C)−

q−1

X

i=1

2dim(Bi) + (7q−6)dim(W|A, B₁, . . . , Bq−1) + (6q−5)dim(X|B₁, . . . , Bq−1, C) +

q−1

X

i=1

(2q)dim(V_i|X,∪^q−1_j=1,j6=iB_j) + (3q−3)dim(Y|W, X) + (4q−3)dim(Z|W, C) + (2q−1)dim(A|Z, V₁, . . . , Vq−1) + (q−1)dim(C|A, Y)

+

q−1

X

i=1

2dim(B_i|B₁, . . . , Bi−1, B_i+1, . . . , Bq−1, Y, Z) +

q−1

X

i=1

dim(C|V_i, B_i) + (5q−4)(dim(A)−dim(A, B₁, . . . , Bq−1, C))

+ (6q−5)(

q−1

X

i=1

dim(B_i) +dim(C))−(q−1)dim(B₁, . . . , Bq−1, C).

5.4 Proof of the Inequality Shown in Equation (5.2)

A

Y

B_i

V_i

B_j

fAX

f_XA

f_AY g_AY f_AV

i f_{V A}

i

f_{B X}

i

f_XB

i

f_XB

j

f_{B X}

j

f_B

iV_i f_{V B}

j

i f_YB

j

f_YB

i X

Figure 5.4: In this figure each circle with a label inside represents a vector subspace. We assume a set of mappings between these subspaces; these mappings are shown with an arrow directed from the domain to the co-domain (note that these mappings are shown for a particular value of iand j, and holds for every value ifiand j where 1≤i, j≤q,j6=i).

fAX :A→X fAY :A→Y

g_AY :A→Y for 1≤i≤q: f_AVi :A→V_i

for 1≤i≤q : fBiX :Bi →X fBiVi :Bi→Vi

f_XA:X →A for 1≤i≤q: f_XBi :X→B_i

for 1≤i≤q : f_{Y Bi} :Y →B_i for 1≤i≤q: f_ViA:V_i→A for 1≤i, j≤q, j 6=i: fViBj :Vi →Bj.

Due to Lemma 8, the following holds:

f_XA+

q

X

i=1

f_XBi =I over a subspace X⁰ of X wherecodim_X(X⁰)≤dim(X|A, B₁, . . . , B_q) (5.97) f_ViA+

q

X

j=1,j6=i

f_ViBj =I over a subspaceV_i⁰ of V_i where

codim_Vi(V_i⁰)≤dim(V_i|A, B₁, . . . , Bi−1, B_i+1, . . . , B_q) (5.98)

q

X

i=1

f_{Y Bi} =I over a subspaceY⁰ of Y wherecodim_Y(Y⁰)≤dim(Y|B₁, . . . , B_q) (5.99) fAX +fAY =I over a subspaceA⁰ of Awhere codimA(A⁰)≤dim(A|X, Y) (5.100)

TH-2118_136102023

q

X

i=1

f_AVi+g_AY =I over a subspaceA⁰⁰ ofA wherecodim_A(A⁰⁰)≤dim(A|V₁, . . . , V_q, Y) (5.101) for 1≤i≤q: fBiX +fBiVi =I over a subspaceB_i⁰ of Bi where

codim_Bi(B_i⁰)≤dim(Bi|X, V_i). (5.102) Consider the following composite functions.

fXAfAX :A→A (5.103)

for 1≤i≤q :f_XBif_AX +f_{Y Bi}f_AY :A→Bi. (5.104) Using equation (5.97), we have:

fXAfAX +

q

X

i=1

fXBifAX =fAX over a subspacef_AX⁻¹(X⁰) of A. (5.105) Using equation (5.99), we have:

q

X

i=1

fY BifAY =fAY over a subspacef_AY⁻¹(Y⁰) of A. (5.106) Consider a subspace A⁰⁰⁰ where

A⁰⁰⁰=f_AX⁻¹(X⁰)∩f_AY⁻¹(Y⁰)∩A⁰. (5.107) Summing the functions shown in equations (5.103) and (5.104), we have

f_XAf_AX+

q

X

i=1

(f_XBif_AX+f_{Y Bi}f_AY)

=f_AX+f_AY =I over a subspaceA⁰⁰⁰ of A. (5.108) So, due to Lemma 9 there exists a subspace ¯A of A⁰⁰⁰ over which:

f_XAf_AX −I = 0 (5.109)

for 1≤i≤q :f_XBif_AX +f_{Y Bi}f_AY = 0. (5.110) wherecodimA⁰⁰⁰( ¯A)≤dim(A) +

q

X

i=1

dim(Bi)−dim(A, B1, . . . , Bq). (5.111) Applying Lemma 6 to equation (5.107), we get:

codim_A(A⁰⁰⁰)≤codim_A(f_AX⁻¹(X⁰)) +codim_A(f_AY⁻¹(Y⁰)) +codim_A(A⁰). (5.112)

5.4 Proof of the Inequality Shown in Equation (5.2)

Applying Lemma 7 to equation (5.112), we get:

codim_A(A⁰⁰⁰)≤codim_X(X⁰) +codim_Y(Y⁰) +codim_A(A⁰). (5.113) From equations (5.113) and (5.111), we have:

codimA( ¯A) =codimA(A⁰⁰⁰) +codim_A⁰⁰⁰( ¯A)≤codimX(X⁰) +codimY(Y⁰) +codimA(A⁰) +dim(A) +

q

X

i=1

dim(B_i)−dim(A, B₁, . . . , B_q). (5.114) Now, for each value of 1≤i≤q, consider the following composite functions.

fXAfBiX +fViAfBiVi :Bi→A (5.115)

f_XBif_BiX :B_i →B_i (5.116)

for 1≤j≤q, j6=i: f_XBjf_BiX +f_ViBjf_BiVi :B_i→B_j. (5.117) Using equation (5.97), we have:

f_XAf_BiX+

q

X

j=1

f_XBjf_BiX =f_BiX over a subspacef_B⁻¹

iX(X⁰) ofB_i. (5.118) Using equation (5.98), we have:

fViAfBiVi+

q

X

j=1,j6=i

fViBjfBiVi =fBiVi over a subspacef_B⁻¹

iVi(V_i⁰) ofVi. (5.119) Consider the following subspace

B⁰⁰_i =f_B⁻¹

iX(X⁰)∩f_B⁻¹

iVi(V_i⁰)∩B_i⁰. (5.120) Summing the functions shown in equations (5.115)-(5.117) overB_i⁰⁰, we get:

f_XAf_BiX +f_ViAf_BiVi +f_XBif_BiX+

q

X

j=1,j6=i

(f_XBjf_BiX+f_ViBjf_BiVi)

=f_XAf_BiX +

q

X

j=1

f_XBjf_BiX +f_ViAf_BiVi+

q

X

j=1,j6=i

f_ViBjf_BiVi

=fBiX+fBiVi [from equations (5.118) and (5.119)]

=I.

TH-2118_136102023

So according to Lemma 9 there exists a subspace ¯Bi over which the following identities hold:

f_XAf_BiX+f_ViAf_BiVi = 0 (5.121)

f_XBif_BiX −I = 0 (5.122)

for 1≤j≤q, j6=i: fXBjfBiX +fViBjfBiVi = 0. (5.123) wherecodim_B⁰⁰

i( ¯B_i)≤dim(A) +

q

X

i=1

dim(B_i)−dim(A, B₁, . . . , B_q). (5.124) Applying Lemma 6 to equation (5.120), we have:

codimBi(B_i⁰⁰) =codimBi(f_B⁻¹

iX(X⁰)) +codimBi(f_B⁻¹

iVi(V_i⁰)) +codimBi(B_i⁰). (5.125) Applying Lemma 7 to equation (5.125), we have:

codim_Bi(B_i⁰⁰) =codim_X(X⁰) +codim_Vi(V_i⁰) +codim_Bi(B_i⁰). (5.126) From equations (5.124) and (5.126), we have:

codim_Bi( ¯B_i) =codim_Bi(B_i⁰⁰) +codim_B⁰⁰

i( ¯B_i)≤codim_X(X⁰) +codim_Vi(V_i⁰) +codim_Bi(B_i⁰) +dim(A) +

q

X

j=1

dim(B_j)−dim(A, B₁, . . . , B_q). (5.127) Now consider the following composite functions:

q

X

i=1

fViAfAVi :A→A (5.128)

for 1≤i≤q :

q

X

j=1,j6=i

fVjBifAVj+fY BigAY :A→Bi. (5.129) Using equation (5.98), we have:

for 1≤i≤q: fViAfAVi+

q

X

j=1,j6=i

fViBjfAVi =fAVi over a subspace f_AV⁻¹

i(V_i⁰) ofA. (5.130) Using equation (5.99), we have:

q

X

i=1

fY BigAY =gAY over a subspaceg_AY⁻¹(Y⁰) of A. (5.131) Consider the following subspace

A⁰⁰⁰⁰= (∩^q_i=1f_AV⁻¹

i(V_i⁰))∩g_AY⁻¹(Y⁰)∩A⁰⁰. (5.132)

5.4 Proof of the Inequality Shown in Equation (5.2)

Summing the functions shown in equations (5.128) and (5.129) overA⁰⁰⁰⁰, we get:

q

X

i=1

f_ViAf_AVi+

q

X

i=1

(

q

X

j=1,j6=i

f_VjBif_AVj+f_{Y Bi}g_AY)

=

q

X

i=1

fViAfAVi+

q

X

i=1 q

X

j=1,j6=i

fVjBifAVj+

q

X

i=1

fY BigAY

=

q

X

i=1

f_ViAf_AVi+

q

X

j=1 q

X

i=1,i6=j

f_VjBif_AVj+

q

X

i=1

f_{Y Bi}g_AY

=

q

X

i=1

fViAfAVi+

q

X

i=1 q

X

j=1,j6=i

fViBjfAVi+

q

X

i=1

fY BigAY

=

q

X

i=1

f_AVi+g_AY Applying equations (5.130) and (5.131)

=I. Using equation (5.101) (5.133)

Then, due to Lemma 9, over a subspace ˆA of A⁰⁰⁰⁰, we have:

q

X

i=1

f_ViAf_AVi−I = 0 (5.134)

for 1≤i≤q:

q

X

j=1,j6=i

fVjBifAVj+fY BigAY = 0. (5.135) where,

codim_A⁰⁰⁰⁰( ˆA)≤dim(A) +

q

X

i=1

dim(Bi)−dim(A, B1, . . . , Bq). (5.136) Applying Lemma 6 on equation (5.132), we get:

codim_A(A⁰⁰⁰⁰) =

q

X

i=1

codim_A(f_AV⁻¹

i(V_i⁰)) +codim_A(g⁻¹_AY(Y⁰)) +codim_A(A⁰⁰). (5.137) Applying Lemma 7 on equation (5.137), we get:

codimA(A⁰⁰⁰⁰) =

q

X

i=1

codimVi(V_i⁰) +codimY(Y⁰) +codimA(A⁰⁰). (5.138) Using equations (5.138) and (5.136), we have:

codimA( ˆA) =codimA(A⁰⁰⁰⁰) +codim_A⁰⁰⁰⁰( ˆA)≤

q

X

i=1

codimVi(V_i⁰) +codimY(Y⁰) +codimA(A⁰⁰) +dim(A) +

q

X

i=1

dim(Bi)−dim(A, B1, . . . , Bq). (5.139)

TH-2118_136102023

Consider the following subspaces.

A^∗ =f_AX( ¯A) (5.140)

A^∗∗=A^∗∩B₁^∗ (5.141)

A^∗∗∗=fXA(A^∗∗) (5.142)

for 1≤i≤q : B_i^∗=f_BiX( ¯B_i) (5.143) B₁^∗∗=B^∗₁∩B₂^∗∩ · · · ∩B_q^∗ (5.144) for 2≤i≤q : B_i^∗∗=B₁^∗∩B_i^∗ (5.145) for 1≤i≤q : B_i^∗∗∗=f_XBi(B_i^∗∗). (5.146) From equation (5.109) and (5.140) we know that fXA is one-to-one overA^∗. Now from (5.140) , we get f_XA(A^∗) = ¯A, which impliesf_XA(A^∗∗)⊆A¯ (as due to (5.141)A^∗∗⊆A^∗). This implies A^∗∗∗⊆A¯ (from (5.142)). Then, f_XAf_AX(A^∗∗∗) = A^∗∗∗ = f_XA(A^∗∗), and so we must have A^∗∗ = f_AX(A^∗∗∗).

With similar reasoning, from equation (5.122) and (5.143)-(5.146) we have: B_i^∗∗ = f_BiX(B_i^∗∗∗) for 1≤i≤q. [This is done in the following way. From equation (5.122) and (5.143) we know thatf_XBi is one-to-one overB_i^∗. Then from (5.143): f_XBi(B_i^∗) = ¯B_i which impliesf_XBi(B^∗∗_i )⊆B¯_i (due to (5.144) and (5.145)). This implies B_i^∗∗∗ ⊆ B¯i (from (5.146)). Then, fXBifBiX(B_i^∗∗∗) = B_i^∗∗∗ = fXBi(B_i^∗∗), and so we must haveB_i^∗∗=fBiX(B_i^∗∗∗).]

Let us define the following subspaces:

Sa ={a∈A|g_AY(a)∈f_AY(A^∗∗∗)} (5.147)

for 1≤i≤q: S_i ={a∈A|f_AVi(a)∈f_BiVi(B_i^∗∗∗)} (5.148) S = ˆA∩Sa∩S1∩S2∩ · · · ∩Sq. (5.149) Let ˆa∈ S. Then f_AVi(ˆa) = f_BiVi(b_i) for some b_i ∈ B_i^∗∗∗ where 1≤i≤q. Alsog_AY(ˆa) = f_AY(a) for somea∈A^∗∗∗. So from equations (5.134) and (5.135) respectively, we have:

q

X

i=1

f_ViAf_BiVi(b_i) = ˆa (5.150)

and for 1≤i≤q:

q

X

j=1,j6=i

f_VjBif_BjVj(b_j) +f_{Y Bi}f_AY(a) = 0. (5.151)

5.4 Proof of the Inequality Shown in Equation (5.2)

Summing equation (5.121) for 1≤i≤q, we have:

q

X

i=1

(f_XAf_BiX +f_ViAf_BiVi)(b_i) = 0 (5.152) Substituting Pq

i=1f_ViAf_BiVi(b_i) from equation (5.150) in equation (5.152), we have:

q

X

i=1

f_XAf_BiX(b_i) =−ˆa (5.153)

interchanging iand j in equation (5.123), and then summing for 1≤j≤q, j6=i, we have:

for 1≤i≤q:

q

X

j=1,j6=i

(f_XBif_BjX +f_VjBif_BjVj)(b_j) = 0 (5.154) Substituting Pq

j=1,j6=if_VjBif_BjVj(b_j) from equation (5.151) in equation (5.154), we have:

for 1≤i≤q:

q

X

j=1,j6=i

fXBifBjX(bj)−fY BifAY(a) = 0 (5.155)

Substituting f_{Y Bi}f_AY(a) from equation (5.110), we have:

for 1≤i≤q:

q

X

j=1,j6=i

f_XBif_BjX(b_j) +f_XBif_AX(a) = 0 (5.156) Fori= 1, from equation (5.156), we get:

q

X

j=2

f_XB1f_BjX(b_j) +f_XB1f_AX(a) = 0 or,f_XB1(

q

X

j=2

f_BjX(b_j) +f_AX(a)) = 0

SincefBjX(bj)∈(B₁^∗∩B^∗_j) for 2≤j ≤q; andfAX(a)∈(A^∗∩B^∗₁); and asfXB1 is invertible overB₁^∗, we have:

q

X

j=2

fBjX(bj) +fAX(a) = 0. (5.157)

For 2≤i≤q, from equation (5.156), we get:

f_XBi(

q

X

j=1,j6=i

f_BjX(b_j) +f_AX(a)) = 0

or,fXBi(fB1X(b1)−fBiX(bi) +

q

X

j=2

fBjX(bj) +fAX(a)) = 0 or,f_XBi(f_B1X(b₁)−f_BiX(b_i)) = 0 [using equation (5.157)]

or, (fB1X(b1)−fBiX(bi)) = 0 [Since fB1X(b1)∈B_i^∗ and fXBi is invertible over B_i^∗ ]

TH-2118_136102023

or, fB1X(b1) =fBiX(bi). (5.158) Substituting equation (5.158) in equation (5.153), we get:

q

X

i=1

f_XAf_B1X(b₁) =−ˆa or, qf_XAf_B1X(b₁) =−ˆa

Now, if the characteristic of the finite field divides q, we must haveq = 0. So, −ˆa= 0.

Since this is true for any arbitrary ˆa∈S, we must have S={0}, which impliesdim(S) = 0.

Now, dim(A) =dim(A)−dim(S) =codim_A(S) =codim_A( ˆA∩S_a∩S₁∩S₂∩ · · · ∩S_q)

≤codimA( ˆA) +codimA(Sa) +

q

X

i=1

codimA(Si) [applying Lemma 6]. (5.159) Hence, from (5.147) we have:

codim_A(S_a) =codim_A(g⁻¹_AY(f_AY(A^∗∗∗)))≤codim_Y(f_AY(A^∗∗∗)) [from Lemma 7]

or, codim_A(S_a)≤dim(Y)−dim(f_AY(A^∗∗∗)) (5.160)

Since fAX(A^∗∗∗) is a subspace of B₁^∗( due to (5.141) and that A^∗∗=fAX(A^∗∗∗)), fXB1fAX is invertible overA^∗∗∗.So from equation (5.110): f_{Y B1}f_AY is invertible over A^∗∗∗, and hence

fAY is invertible over A^∗∗∗.Then,dim(fAY(A^∗∗∗)) =dim(A^∗∗∗).Hence from eqn. (5.160) we have:

codim_A(Sa)≤dim(Y)−dim(A^∗∗∗) =dim(Y)−dim(f_XA(A^∗∗))

Now, A^∗∗ is a subspace off_AX( ¯A), and over f_AX( ¯A),f_XA is invertible because of eqn. (5.109).

So,codimA(Sa)≤dim(Y)−dim(A^∗∗) =dim(Y)−dim(A^∗∩B^∗₁)

= dim(Y) +codim_X(A^∗∩B₁^∗)−dim(X) or, codim_A(S_a)≤dim(Y) +codim_X(A^∗) +codim_X(B₁^∗)−dim(X) [from Lemma 6]

or, codimA(Sa)≤dim(Y) +dim(X)−dim(A^∗) +dim(X)−dim(B₁^∗)−dim(X) or, codim_A(S_a)≤dim(Y) +dim(X)−dim(f_AX( ¯A))−dim(f_B1X( ¯B₁))

or, codim_A(S_a)≤dim(Y) +dim(X)−dim( ¯A)−dim( ¯B₁)

or, codimA(Sa)≤dim(Y) +dim(X) +codimA( ¯A) +codimB1( ¯B1)−dim(A)−dim(B1). (5.161)

5.4 Proof of the Inequality Shown in Equation (5.2)

for 1≤i≤q, from (5.148) we have:

codim_A(S_i) =codim_A(f_AV⁻¹

i(f_BiVi(B_i^∗∗∗)))≤codim_Vi(f_BiVi(B_i^∗∗∗)) [from Lemma 7]

or, codim_A(S_i)≤dim(V_i)−dim(f_BiVi(B_i^∗∗∗))

Since fBiX(B_i^∗∗∗) is a subspace of B₁^∗, fXB1fBiX is invertible over B^∗∗∗_i ; from equation (5.123) f_ViB1f_BiVi is invertible over B_i^∗∗∗, and hence f_BiVi must be invertible overB_i^∗∗∗.So,

codimA(Si)≤dim(Vi)−dim(Bi∗∗∗

) =dim(Vi)−dim(fXBi(Bi∗∗

)) (5.162)

Now, B^∗∗_i is a subspace offBiX( ¯Bi), and overfBiX( ¯Bi): fXBi is invertible from eq. (5.122).

So, for 2≤i≤q we have:

codimA(Si)≤dim(Vi)−dim(Bi∗∗

) =dim(Vi)−dim(Bi∗∩B^∗₁)

= dim(V_i) +codim_X(B_i^∗∩B₁^∗)−dim(X)

≤dim(V_i) +codim_X(B_i^∗) +codim_X(B₁^∗)−dim(X) [using Lemma 6]

= dim(Vi) +dim(X)−dim(B^∗_i) +dim(X)−dim(B₁^∗)−dim(X)

= dim(V_i) +dim(X)−dim(f_BiX( ¯B_i))−dim(f_B1X( ¯B₁))

= dim(V_i) +dim(X)−dim( ¯B_i)−dim( ¯B₁) [using equation (5.122)]

= dim(Vi) +dim(X) +codimBi( ¯Bi) +codimB1( ¯B1)−dim(Bi)−dim(B1) (5.163) fori= 1 form equation (5.162) we have:

codimA(S1)≤dim(V1)−dim(B1∗∗

) =dim(V1)−dim(B1∗∩B₂^∗∩ · · · ∩B^∗_q)

= dim(V1) +codim_X(B1∗∩B₂^∗∩ · · · ∩B_q^∗)−dim(X)

≤dim(V1) +

q

X

i=1

codimX(Bi∗)−dim(X) [from Lemma 6]

= dim(V1) + (q)dim(X)−

q

X

i=1

dim(Bi∗

)−dim(X)

= dim(V1) + (q−1)dim(X)−

q

X

i=1

dim(fBiX( ¯Bi))

= dim(V₁) + (q−1)dim(X)−

q

X

i=1

dim( ¯B_i) [using equation (5.122)]

= dim(V₁) + (q−1)dim(X) +

q

X

i=1

codim_Bi( ¯B_i)−

q

X

i=1

dim(B_i). (5.164)

TH-2118_136102023

So, substituting equations (5.161), (5.163), and (5.164) in equation (5.159), we have:

dim(A)≤codim_A( ˆA) +codim_A(S_a) +

q

X

i=1

codim_A(S_i)

or, dim(A)≤codim_A( ˆA) +dim(Y) +dim(X) +codim_A( ¯A) +codim_B1( ¯B₁)−dim(A)

−dim(B1) +dim(V1) + (q−1)dim(X) +

q

X

i=1

codimBi( ¯Bi)−

q

X

i=1

dim(Bi) +

q

X

i=2

dim(Vi) + (q−1)dim(X) +

q

X

i=2

codimBi( ¯Bi) + (q−1)codimB1( ¯B1)−

q

X

i=2

dim(Bi)−(q−1)dim(B1) or, dim(A)≤codimA( ˆA) +dim(Y) + (2q−1)dim(X) +codimA( ¯A)−dim(A)

−(q+ 1)dim(B₁) +

q

X

i=1

dim(V_i) +

q

X

i=2

2codim_Bi( ¯B_i) + (q+ 1)codim_B1( ¯B₁)−

q

X

i=2

2dim(B_i) Substituting values from equations (5.139), (5.114), and (5.127), we get:

dim(A)≤

q

X

i=1

codimVi(V_i⁰) +codimY(Y⁰) +codimA(A⁰⁰) +dim(A) +

q

X

i=1

dim(Bi)

−dim(A, B1, . . . , Bq) +dim(Y) + (2q−1)dim(X) +codimX(X⁰) +codimY(Y⁰) +codimA(A⁰) +dim(A) +

q

X

i=1

dim(B_i)−dim(A, B₁, . . . , B_q)−dim(A)−(q+ 1)dim(B₁) +

q

X

i=1

dim(V_i) +

q

X

i=2

2(codim_X(X⁰)+codim_Vi(V_i⁰) +codim_Bi(B_i⁰) +dim(A) +

q

X

j=1

dim(Bj)−dim(A, B₁, .., Bq)) + (q+ 1)(codim_X(X⁰) +codim_V1(V₁⁰) +codim_B1(B₁⁰) +dim(A) +

q

X

j=1

dim(B_j)

−dim(A, B₁, . . . , B_q))−

q

X

i=2

2dim(B_i) or, dim(A)≤(q+ 2)codim_V1(V₁⁰) +

q

X

i=2

3codim_Vi(V_i⁰) + 2codim_Y(Y⁰) +codim_A(A⁰⁰) +dim(Y) + (2q−1)dim(X) + (3q)codim_X(X⁰) +codim_A(A⁰)−dim(A)−(q+ 1)dim(B₁) +

q

X

i=1

dim(V_i) +

q

X

i=2

2codimBi(B_i⁰) + (q+ 1)codimB1(B₁⁰)−

q

X

i=2

2dim(Bi) + (3q+ 1)(dim(A) +

q

X

j=1

dim(Bj)−dim(A, B1, . . . , Bq))

Substituting values from equations (5.97), (5.99), (5.98), (5.100), (5.102) and (5.101), we get:

or, dim(A)≤(q+ 2)dim(V₁|A, B₂, . . . , B_q) +

q

X

i=2

3dim(V_i|A, B₁, . . . , Bi−1, B_i+1, . . . , B_q) + 2dim(Y|B₁, . . . , B_q) +dim(A|V₁, . . . , V_q, Y) +dim(Y) + (2q−1)dim(X)

+ (3q)dim(X|A, B₁, . . . , Bq) +dim(A|X, Y)−dim(A)−(q+ 1)dim(B1) +

q

X

i=1

dim(Vi)