let V be set ; :: thesis: for G being Graph
for pe being FinSequence of the carrier' of G st not vertices pe c= V holds
ex i being Element of NAT ex q, r being FinSequence of the carrier' of G st
( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V )

let G be Graph; :: thesis: for pe being FinSequence of the carrier' of G st not vertices pe c= V holds
ex i being Element of NAT ex q, r being FinSequence of the carrier' of G st
( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V )

let pe be FinSequence of the carrier' of G; :: thesis: ( not vertices pe c= V implies ex i being Element of NAT ex q, r being FinSequence of the carrier' of G st
( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V ) )

defpred S1[ Nat] means ( $1 in dom pe & not vertices (pe /. $1) c= V );
assume not vertices pe c= V ; :: thesis: ex i being Element of NAT ex q, r being FinSequence of the carrier' of G st
( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V )

then A1: ex i being Nat st S1[i] by Th17;
consider k being Nat such that
A2: ( S1[k] & ( for n being Nat st S1[n] holds
k <= n ) ) from NAT_1:sch 5(A1);
k <= len pe by A2, FINSEQ_3:25;
then consider j being Nat such that
A3: len pe = k + j by NAT_1:10;
reconsider j = j as Element of NAT by ORDINAL1:def 12;
1 <= k by A2, FINSEQ_3:25;
then consider i being Nat such that
A4: k = 1 + i by NAT_1:10;
reconsider i = i as Element of NAT by ORDINAL1:def 12;
len pe = i + (1 + j) by A4, A3;
then consider q, r being FinSequence such that
A5: len q = i and
len r = 1 + j and
A6: pe = q ^ r by FINSEQ_2:22;
reconsider q = q, r = r as FinSequence of the carrier' of G by A6, FINSEQ_1:36;
take i ; :: thesis: ex q, r being FinSequence of the carrier' of G st
( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V )

take q ; :: thesis: ex r being FinSequence of the carrier' of G st
( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V )

take r ; :: thesis: ( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V & len q = i & pe = q ^ r & vertices q c= V )
thus ( i + 1 <= len pe & not vertices (pe /. (i + 1)) c= V ) by A2, A4, FINSEQ_3:25; :: thesis: ( len q = i & pe = q ^ r & vertices q c= V )
thus ( len q = i & pe = q ^ r ) by A5, A6; :: thesis: vertices q c= V
now :: thesis: for m being Nat st m in dom q holds
vertices (q /. m) c= V
let m be Nat; :: thesis: ( m in dom q implies vertices (q /. m) c= V )
assume A7: m in dom q ; :: thesis: vertices (q /. m) c= V
then A8: m <= len q by FINSEQ_3:25;
A9: dom q c= dom pe by A6, FINSEQ_1:26;
assume A10: not vertices (q /. m) c= V ; :: thesis: contradiction
q /. m = q . m by A7, PARTFUN1:def 6
.= pe . m by A6, A7, FINSEQ_1:def 7
.= pe /. m by A7, A9, PARTFUN1:def 6 ;
then k <= m by A2, A7, A10, A9;
hence contradiction by A4, A5, A8, NAT_1:13; :: thesis: verum
end;
hence vertices q c= V by Th17; :: thesis: verum