defpred S_{1}[ Nat] means BCS (n,K) is finite-vertices ;

[#] K = [#] V by SIMPLEX0:def 10;

then A1: |.K.| c= [#] K ;

A2: for n being Nat st S_{1}[n] holds

S_{1}[n + 1]
_{1}[ 0 ]
by A1, Th16;

for n being Nat holds S_{1}[n]
from NAT_1:sch 2(A5, A2);

hence BCS (n,K) is finite-vertices ; :: thesis: verum

[#] K = [#] V by SIMPLEX0:def 10;

then A1: |.K.| c= [#] K ;

A2: for n being Nat st S

S

proof

A5:
S
let n be Nat; :: thesis: ( S_{1}[n] implies S_{1}[n + 1] )

assume A3: S_{1}[n]
; :: thesis: S_{1}[n + 1]

[#] (BCS (n,K)) = [#] V by SIMPLEX0:def 10;

then A4: |.(BCS (n,K)).| c= [#] (BCS (n,K)) ;

BCS ((n + 1),K) = BCS (BCS (n,K)) by A1, Th20

.= subdivision ((center_of_mass V),(BCS (n,K))) by A4, Def5 ;

hence S_{1}[n + 1]
by A3; :: thesis: verum

end;assume A3: S

[#] (BCS (n,K)) = [#] V by SIMPLEX0:def 10;

then A4: |.(BCS (n,K)).| c= [#] (BCS (n,K)) ;

BCS ((n + 1),K) = BCS (BCS (n,K)) by A1, Th20

.= subdivision ((center_of_mass V),(BCS (n,K))) by A4, Def5 ;

hence S

for n being Nat holds S

hence BCS (n,K) is finite-vertices ; :: thesis: verum