let n be Nat; :: thesis: (((2 * n) + 1) ^2) mod 8 = 1
defpred S₁[ Nat] means (((2 * $1) + 1) ^2) mod 8 = 1;
A1: S₁[ 0 ] by NAT_D:24;
A2: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A3: S₁[k] ; :: thesis: S₁[k + 1]
thus (((2 * (k + 1)) + 1) ^2) mod 8 = ((((4 * (k ^2)) + (4 * k)) + 1) + (8 * (k + 1))) mod 8
.= (1 + ((8 * (k + 1)) mod 8)) mod 8 by A3, NAT_D:66
.= (1 + 0) mod 8 by NAT_D:13
.= 1 by NAT_D:24 ; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A1, A2);
hence (((2 * n) + 1) ^2) mod 8 = 1 ; :: thesis: verum