let n be Nat; :: thesis: Ackermann (4,n) = (2 |^|^ (n + 3)) - 3
defpred S₁[ Nat] means H₁(4,$1) = (2 |^|^ ($1 + 3)) - 3;
A1: S₁[ 0 ]

proof

A2: 2 |^|^ 3 = exp (2,(exp (2,2))) by ORDINAL5:19;
thus H₁(4, 0 ) = H₁(4 - 1,1) by Def1
.= (2 |^ (1 + 3)) - 3 by Th11
.= (2 |^|^ (0 + 3)) - 3 by A2, Lm1 ; :: thesis: verum

end;

A3: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A4: S₁[k] ; :: thesis: S₁[k + 1]
A5: H₁(4,k) + 3 = 2 |^|^ (k + 3) by A4;
succ (Segm (k + 3)) = Segm ((k + 3) + 1) by NAT_1:38;
then A6: 2 |^|^ (k + 4) = exp (2,(2 |^|^ (k + 3))) by ORDINAL5:14
.= 2 |^ (2 |^|^ (k + 3)) ;
thus H₁(4,k + 1) = H₁(4 - 1,H₁(4,(k + 1) - 1)) by Def1
.= (2 |^|^ ((k + 1) + 3)) - 3 by A5, A6, Th11 ; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A1, A3);
hence Ackermann (4,n) = (2 |^|^ (n + 3)) - 3 ; :: thesis: verum