defpred S1[ Nat, set , set ] means ex i1, i2 being Integer st
( i1 = \$2 & i2 = \$3 & i2 = ((() * i1) + (Table1 (q,c,f,(n -' \$1)))) mod f );
reconsider n1 = n as Element of NAT by ORDINAL1:def 12;
reconsider T = Table1 (q,c,f,n) as Element of INT by INT_1:def 2;
A2: for i being Nat st 1 <= i & i < n1 holds
for x being Element of INT ex y being Element of INT st S1[i,x,y]
proof
let i be Nat; :: thesis: ( 1 <= i & i < n1 implies for x being Element of INT ex y being Element of INT st S1[i,x,y] )
assume that
1 <= i and
i < n1 ; :: thesis: for x being Element of INT ex y being Element of INT st S1[i,x,y]
let x be Element of INT ; :: thesis: ex y being Element of INT st S1[i,x,y]
reconsider x = x as Integer ;
consider y being Integer such that
A3: y = ((() * x) + (Table1 (q,c,f,(n -' i)))) mod f ;
reconsider z = y as Element of INT by INT_1:def 2;
take z ; :: thesis: S1[i,x,z]
thus S1[i,x,z] by A3; :: thesis: verum
end;
consider r being FinSequence of INT such that
A4: ( len r = n1 & ( r . 1 = T or n1 = 0 ) ) and
A5: for i being Nat st 1 <= i & i < n1 holds
S1[i,r . i,r . (i + 1)] from reconsider r = r as Tuple of n, INT by ;
take r ; :: thesis: ( r . 1 = Table1 (q,c,f,n) & ( for i being Nat st 1 <= i & i <= n - 1 holds
ex I1, I2 being Integer st
( I1 = r . i & I2 = r . (i + 1) & I2 = ((() * I1) + (Table1 (q,c,f,(n -' i)))) mod f ) ) )

thus r . 1 = Table1 (q,c,f,n) by A1, A4; :: thesis: for i being Nat st 1 <= i & i <= n - 1 holds
ex I1, I2 being Integer st
( I1 = r . i & I2 = r . (i + 1) & I2 = ((() * I1) + (Table1 (q,c,f,(n -' i)))) mod f )

let i be Nat; :: thesis: ( 1 <= i & i <= n - 1 implies ex I1, I2 being Integer st
( I1 = r . i & I2 = r . (i + 1) & I2 = ((() * I1) + (Table1 (q,c,f,(n -' i)))) mod f ) )

assume A6: ( 1 <= i & i <= n - 1 ) ; :: thesis: ex I1, I2 being Integer st
( I1 = r . i & I2 = r . (i + 1) & I2 = ((() * I1) + (Table1 (q,c,f,(n -' i)))) mod f )

thus ex I1, I2 being Integer st
( I1 = r . i & I2 = r . (i + 1) & I2 = ((() * I1) + (Table1 (q,c,f,(n -' i)))) mod f ) by ; :: thesis: verum