let n be Nat; :: thesis: Triangle n = (n * (n + 1)) / 2
defpred S₁[ Nat] means Triangle $1 = ($1 * ($1 + 1)) / 2;
A1: S₁[ 0 ] ;
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]
Triangle (k + 1) = (Triangle k) + (k + 1) by Th10
.= ((k + 1) * (k + 2)) / 2 by A3 ;
hence S₁[k + 1] ; :: thesis: verum

end;

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