let n be Nat; :: thesis: (n + 1) block n = (n + 1) choose 2
defpred S₁[ Nat] means ($1 + 1) block $1 = ($1 + 1) choose 2;
A1: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A2: S₁[k] ; :: thesis: S₁[k + 1]
set k1 = k + 1;
thus ((k + 1) + 1) block (k + 1) = ((k + 1) * ((k + 1) block (k + 1))) + ((k + 1) block k) by Th46
.= ((k + 1) * 1) + ((k + 1) choose 2) by A2, Th26
.= (k + 1) + (((k + 1) * ((k + 1) - 1)) / 2) by Th51
.= (((k + 1) + 1) * (((k + 1) + 1) - 1)) / 2
.= ((k + 1) + 1) choose 2 by Th51 ; :: thesis: verum

end;

1 block 0 = 0 by Th31;
then A3: S₁[ 0 ] by NEWTON:def 3;
for k being Nat holds S₁[k] from NAT_1:sch 2(A3, A1);
hence (n + 1) block n = (n + 1) choose 2 ; :: thesis: verum