let n be Element of NAT ; :: thesis: (n + 1) block n = (n + 1) choose 2
defpred S₁[ Element of NAT ] means ($1 + 1) block $1 = ($1 + 1) choose 2;
A1: S₁[ 0 ]

proof

( 1 block 0 = 0 & 1 choose 2 = 0 ) by Th31, NEWTON:def 3;
hence S₁[ 0 ] ; :: thesis: verum

end;

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: 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 Th56
.= ((k + 1) * 1) + ((k + 1) choose 2) by A3, Th26
.= (k + 1) + (((k + 1) * ((k + 1) - 1)) / 2) by Th61
.= (((k + 1) + 1) * (((k + 1) + 1) - 1)) / 2
.= ((k + 1) + 1) choose 2 by Th61 ; :: thesis: verum

end;

for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A1, A2);
hence (n + 1) block n = (n + 1) choose 2 ; :: thesis: verum