let n be Nat; :: thesis: Triangle (n + 1) = (Triangle n) + (n + 1)
defpred S₁[ Nat] means (Triangle $1) + ($1 + 1) = Triangle ($1 + 1);
A1: S₁[ 0 ] by FINSEQ_2:50, RVSUM_1:73;
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 S₁[k] ; :: thesis: S₁[k + 1]
reconsider k1 = k as Element of NAT by ORDINAL1:def 12;
(((Triangle (k + 1)) + k) + 1) + 1 = (((Sum ((idseq k1) ^ <*(k1 + 1)*>)) + k) + 1) + 1 by FINSEQ_2:51
.= (Sum ((idseq k1) ^ <*(k1 + 1)*>)) + ((k + 1) + 1)
.= (Sum (idseq (k1 + 1))) + ((k1 + 1) + 1) by FINSEQ_2:51
.= Sum ((idseq (k1 + 1)) ^ <*((k1 + 1) + 1)*>) by RVSUM_1:74
.= Triangle ((k + 1) + 1) by FINSEQ_2:51 ;
hence S₁[k + 1] ; :: thesis: verum

end;

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