consider t being RedSequence of F₃() such that
A4: ( t . 1 = F₁() & t . (len t) = F₂() ) by A2;
defpred S₁[ Nat] means ( $1 in dom t implies P₁[t . $1] );
A5: S₁[ 0 ] by FINSEQ_3:24;

A6: now :: thesis: for i being Nat st S₁[i] holds
S₁[i + 1]

let i be Nat; :: thesis: ( S₁[i] implies S₁[i + 1] )
assume A7: S₁[i] ; :: thesis: S₁[i + 1]
thus S₁[i + 1] :: thesis: verum

proof

assume A8: ( i + 1 in dom t & P₁[t . (i + 1)] ) ; :: thesis: contradiction

per cases ( i = 0 or i > 0 ) by NAT_1:3;

suppose i = 0 ; :: thesis: contradiction

hence contradiction by A1, A4, A8; :: thesis: verum

end;

suppose i > 0 ; :: thesis: contradiction

then consider j being Nat such that
A9: i = j + 1 by NAT_1:6;
( i <= i + 1 & i + 1 <= len t ) by A8, FINSEQ_3:25, NAT_1:11;
then A10: ( 1 <= i & i <= len t & rng t <> {} ) by A9, NAT_1:11, XXREAL_0:2;
then A11: ( i in dom t & 1 in dom t ) by FINSEQ_3:25, FINSEQ_3:32;
[(t . i),(t . (i + 1))] in F₃() by A8, A11, REWRITE1:def 2;
hence contradiction by A3, A7, A8, A11, A4, A10, REWRITE1:17; :: thesis: verum

end;

end;

end;

end;

for i being Nat holds S₁[i] from NAT_1:sch 2(A5, A6);
hence P₁[F₂()] by A4, FINSEQ_5:6; :: thesis: verum