let b, m be FinSequence of INT ; :: thesis: ( len b = len m & ( for i being Nat st i in Seg (len b) holds
b . i <> 0 ) & m . 1 = 1 implies for k being Element of NAT st 1 <= k & k <= (len b) - 1 & ( for i being Nat st 1 <= i & i <= k holds
m . (i + 1) = (m . i) * (b . i) ) holds
m . (k + 1) <> 0 )
assume len b = len m ; :: thesis: ( ex i being Nat st
( i in Seg (len b) & not b . i <> 0 ) or not m . 1 = 1 or for k being Element of NAT st 1 <= k & k <= (len b) - 1 & ( for i being Nat st 1 <= i & i <= k holds
m . (i + 1) = (m . i) * (b . i) ) holds
m . (k + 1) <> 0 )
assume A2: ( ( for i being Nat st i in Seg (len b) holds
b . i <> 0 ) & m . 1 = 1 ) ; :: thesis: for k being Element of NAT st 1 <= k & k <= (len b) - 1 & ( for i being Nat st 1 <= i & i <= k holds
m . (i + 1) = (m . i) * (b . i) ) holds
m . (k + 1) <> 0
defpred S₁[ Nat] means ( 1 <= $1 & $1 <= (len b) - 1 & ( for i being Nat st 1 <= i & i <= $1 holds
m . (i + 1) = (m . i) * (b . i) ) implies m . ($1 + 1) <> 0 );
reconsider I0 = 0 as Element of NAT ;
P0: S₁[ 0 ] ;
P1: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume P11: S₁[k] ; :: thesis: S₁[k + 1]
assume P12: ( 1 <= k + 1 & k + 1 <= (len b) - 1 & ( for i being Nat st 1 <= i & i <= k + 1 holds
m . (i + 1) = (m . i) * (b . i) ) ) ; :: thesis: m . ((k + 1) + 1) <> 0
P14: k <= k + 1 by NAT_1:12;

per cases ( k = 0 or k <> 0 ) ;

suppose P15: k = 0 ; :: thesis: m . ((k + 1) + 1) <> 0

P16: m . ((k + 1) + 1) = (m . 1) * (b . 1) by P12, P15
.= b . 1 by A2 ;
((len b) - 1) + 0 <= ((len b) - 1) + 1 by XREAL_1:7;
then k + 1 <= len b by P12, XXREAL_0:2;
then ( 1 <= 1 & 1 <= len b ) by XXREAL_0:2, P12;
then 1 in Seg (len b) ;
hence m . ((k + 1) + 1) <> 0 by P16, A2; :: thesis: verum

end;

suppose P19: k <> 0 ; :: thesis: m . ((k + 1) + 1) <> 0

P20: now :: thesis: for i being Nat st 1 <= i & i <= k holds
m . (i + 1) = (m . i) * (b . i)

let i be Nat; :: thesis: ( 1 <= i & i <= k implies m . (i + 1) = (m . i) * (b . i) )
assume ( 1 <= i & i <= k ) ; :: thesis: m . (i + 1) = (m . i) * (b . i)
then ( 1 <= i & i <= k + 1 ) by NAT_1:12;
hence m . (i + 1) = (m . i) * (b . i) by P12; :: thesis: verum

end;

thus m . ((k + 1) + 1) <> 0 :: thesis: verum

proof

XX6: (k + 1) + 1 <= ((len b) - 1) + 1 by P12, XREAL_1:6;
XX1: k + 1 <= (k + 1) + 1 by NAT_1:12;
XX2: 1 <= k + 1 by NAT_1:12;
k + 1 <= len b by XX1, XX6, XXREAL_0:2;
then k + 1 in Seg (len b) by XX2;
then P23: b . (k + 1) <> 0 by A2;
m . ((k + 1) + 1) = (m . (k + 1)) * (b . (k + 1)) by P12;
hence m . ((k + 1) + 1) <> 0 by P23, P20, P11, P12, P14, XXREAL_0:2, NAT_1:14, P19, XCMPLX_1:6; :: thesis: verum

end;

end;

end;

end;

for k being Element of NAT holds S₁[k] from NAT_1:sch 1(P0, P1);
hence for k being Element of NAT st 1 <= k & k <= (len b) - 1 & ( for i being Nat st 1 <= i & i <= k holds
m . (i + 1) = (m . i) * (b . i) ) holds
m . (k + 1) <> 0 ; :: thesis: verum