defpred S1[ Nat] means P1[$1];
assume A2:
for k being Nat holds
( not P1[k] or ex n being Nat st
( P1[n] & not k <= n ) )
; contradiction
A3:
for k being Nat st ( for n being Nat st n < k holds
S1[n] ) holds
S1[k]
proof
let k be
Nat;
( ( for n being Nat st n < k holds
S1[n] ) implies S1[k] )
assume A4:
for
n being
Nat st
n < k holds
not
P1[
n]
;
S1[k]
( ( for
n being
Nat holds
( not
P1[
n] or
k <= n ) ) implies not
P1[
k] )
by A2;
hence
S1[
k]
by A4;
verum
end;
for k being Nat holds S1[k]
from NAT_1:sch 4(A3);
hence
contradiction
by A1; verum