defpred S₁[ Element of NAT , set , set ] means ( ( $2 is LexBFS:Labeling of G implies ex L being LexBFS:Labeling of G st
( $2 = L & $3 = LexBFS:Step L ) ) & ( $2 is not LexBFS:Labeling of G implies $3 = $2 ) );

now

let n be Element of NAT ; :: thesis: for x being set ex y being set st S₁[n,x,y]
let x be set ; :: thesis: ex y being set st S₁[n,x,y]

now

per cases ( x is LexBFS:Labeling of G or not x is LexBFS:Labeling of G ) ;

suppose x is LexBFS:Labeling of G ; :: thesis: ex y being set st S₁[n,x,y]

then reconsider L = x as LexBFS:Labeling of G ;
LexBFS:Step L = LexBFS:Step L ;
hence ex y being set st S₁[n,x,y] ; :: thesis: verum

end;

suppose x is not LexBFS:Labeling of G ; :: thesis: ex y being set st S₁[n,x,y]

hence ex y being set st S₁[n,x,y] ; :: thesis: verum

end;

end;

end;

hence ex y being set st S₁[n,x,y] ; :: thesis: verum

end;

then A1: for n being Element of NAT
for x being set ex y being set st S₁[n,x,y] ;
consider IT being Function such that
A2: ( dom IT = NAT & IT . 0 = LexBFS:Init G & ( for n being Element of NAT holds S₁[n,IT . n,IT . (n + 1)] ) ) from RECDEF_1:sch 1(A1);
reconsider IT = IT as ManySortedSet of NAT by A2, PARTFUN1:def 4, RELAT_1:def 18;
defpred S₂[ Nat] means IT . $1 is LexBFS:Labeling of G;

A3: now

let n be Nat; :: thesis: ( S₂[n] implies S₂[n + 1] )
assume A4: S₂[n] ; :: thesis: S₂[n + 1]
n in NAT by ORDINAL1:def 13;
then ex Gn being LexBFS:Labeling of G st
( IT . n = Gn & IT . (n + 1) = LexBFS:Step Gn ) by A2, A4;
hence S₂[n + 1] ; :: thesis: verum

end;

A5: S₂[ 0 ] by A2;
for n being Nat holds S₂[n] from NAT_1:sch 2(A5, A3);
then reconsider IT = IT as LexBFS:LabelingSeq of G by Def15;
take IT ; :: thesis: ( IT . 0 = LexBFS:Init G & ( for n being Nat holds IT . (n + 1) = LexBFS:Step (IT . n) ) )
thus IT . 0 = LexBFS:Init G by A2; :: thesis: for n being Nat holds IT . (n + 1) = LexBFS:Step (IT . n)
let n be Nat; :: thesis: IT . (n + 1) = LexBFS:Step (IT . n)
n in NAT by ORDINAL1:def 13;
then ex Gn being LexBFS:Labeling of G st
( IT . n = Gn & IT . (n + 1) = LexBFS:Step Gn ) by A2;
hence IT . (n + 1) = LexBFS:Step (IT . n) ; :: thesis: verum