defpred S₁[ Element of NAT ] means ( 0 in dyadic $1 & 1 in dyadic $1 );
A1: S₁[ 0 ] by Th6, TARSKI:def 2;
A2: 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 A3: ( 0 in dyadic k & 1 in dyadic k ) ; :: thesis: S₁[k + 1]
dyadic k c= dyadic (k + 1) by Th11;
hence S₁[k + 1] by A3; :: thesis: verum

end;

for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A1, A2);
hence for n being Element of NAT holds
( 0 in dyadic n & 1 in dyadic n ) ; :: thesis: verum