set X = { k where k is Element of NAT : k > 1 } ;
assume A1: { k where k is Element of NAT : k > 1 } is finite ; :: thesis: contradiction
A2: { k where k is Element of NAT : k > 1 } c= NAT

proof

let x be set ; :: according to TARSKI:def 3 :: thesis: ( not x in { k where k is Element of NAT : k > 1 } or x in NAT )
assume x in { k where k is Element of NAT : k > 1 } ; :: thesis: x in NAT
then ex k being Element of NAT st
( x = k & k > 1 ) ;
hence x in NAT ; :: thesis: verum

end;

2 in { k where k is Element of NAT : k > 1 } ;
then reconsider X = { k where k is Element of NAT : k > 1 } as non empty finite Subset of NAT by A1, A2;
set m = max X;
max X in X by XXREAL_2:def 8;
then consider k being Element of NAT such that
A3: max X = k and
A4: k > 1 ;
k + 1 > k + 0 by XREAL_1:10;
then k + 1 > 1 by A4, XXREAL_0:2;
then (max X) + 1 in X by A3;
then (max X) + 1 <= (max X) + 0 by XXREAL_2:def 8;
hence contradiction by XREAL_1:10; :: thesis: verum