let n be Nat; :: thesis: Ackermann (2,n) = (2 * n) + 3
defpred S₁[ Nat] means H₁(2,$1) = (2 * $1) + 3;
A1: S₁[ 0 ]

proof

thus H₁(2, 0 ) = H₁(2 - 1,1) by Def1
.= 2 + 1 by Th9
.= (2 * 0) + 3 ; :: thesis: verum

end;

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 H₁(2,k + 1) = H₁(2 - 1,H₁(2,(k + 1) - 1)) by Def1
.= H₁(2,k) + 2 by Th9
.= (2 * (k + 1)) + 3 by A3 ; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A1, A2);
hence Ackermann (2,n) = (2 * n) + 3 ; :: thesis: verum