defpred S₁[ Nat] means F₃() . $1 <> F₄() . $1;
assume F₃() <> F₄() ; :: thesis: contradiction
then ex x being object st
( x in NAT & F₃() . x <> F₄() . x ) by A1, A4, FUNCT_1:2;
then A7: ex k being Nat st S₁[k] ;
consider m being Nat such that
A8: S₁[m] and
A9: for n being Nat st S₁[n] holds
m <= n from NAT_1:sch 5(A7);

now :: thesis: ( m <> 0 implies not m <> 1 )

set k = m -' 2;
A10: ( F₃() . ((m -' 2) + 2) = F₅((m -' 2),(F₃() . (m -' 2)),(F₃() . ((m -' 2) + 1))) & F₄() . ((m -' 2) + 2) = F₅((m -' 2),(F₄() . (m -' 2)),(F₄() . ((m -' 2) + 1))) ) by A3, A6;
assume ( m <> 0 & m <> 1 ) ; :: thesis: contradiction
then 1 < m by NAT_1:25;
then 1 + 1 <= m by NAT_1:13;
then A11: m = (m -' 2) + 2 by XREAL_1:235;
then A12: F₃() . ((m -' 2) + 1) = F₄() . ((m -' 2) + 1) by A9, XREAL_1:6;
(m -' 2) + 0 < m by A11, XREAL_1:6;
hence contradiction by A8, A9, A11, A12, A10; :: thesis: verum

end;

hence contradiction by A2, A5, A8; :: thesis: verum