defpred S₁[ set , set ] means ex s, mm being Integer st
( mm = (Product m) / (m . $1) & s * mm,1 are_congruent_mod m . $1 & $2 = s * ((Product m) / (m . $1)) );
A1: for k being Nat st k in Seg (len m) holds
ex x being Element of INT st S₁[k,x]

proof

let k be Nat; :: thesis: ( k in Seg (len m) implies ex x being Element of INT st S₁[k,x] )
assume k in Seg (len m) ; :: thesis: ex x being Element of INT st S₁[k,x]
then A2: k in dom m by FINSEQ_1:def 3;
reconsider mm = (Product m) / (m . k) as Integer ;
consider s being Integer such that
A3: s * mm,1 are_congruent_mod m . k by A2, Th22, Th25;
set x = s * mm;
reconsider x = s * mm as Element of INT by INT_1:def 2;
take x ; :: thesis: S₁[k,x]
take s ; :: thesis: ex mm being Integer st
( mm = (Product m) / (m . k) & s * mm,1 are_congruent_mod m . k & x = s * ((Product m) / (m . k)) )
take mm ; :: thesis: ( mm = (Product m) / (m . k) & s * mm,1 are_congruent_mod m . k & x = s * ((Product m) / (m . k)) )
thus ( mm = (Product m) / (m . k) & s * mm,1 are_congruent_mod m . k & x = s * ((Product m) / (m . k)) ) by A3; :: thesis: verum

end;

consider p being FinSequence of INT such that
A4: ( dom p = Seg (len m) & ( for k being Nat st k in Seg (len m) holds
S₁[k,p . k] ) ) from FINSEQ_1:sch 5(A1);
take p ; :: thesis: ( len p = len m & ( for i being Nat st i in dom p holds
ex s, mm being Integer st
( mm = (Product m) / (m . i) & s * mm,1 are_congruent_mod m . i & p . i = s * ((Product m) / (m . i)) ) ) )
thus ( len p = len m & ( for i being Nat st i in dom p holds
ex s, mm being Integer st
( mm = (Product m) / (m . i) & s * mm,1 are_congruent_mod m . i & p . i = s * ((Product m) / (m . i)) ) ) ) by A4, FINSEQ_1:def 3; :: thesis: verum