defpred S₁[ Nat] means F₅() . $1 <> F₆() . $1;
assume F₅() <> F₆() ; :: thesis: contradiction
then ex x being object st
( x in NAT & F₅() . x <> F₆() . 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: not m <> 0 & ... & m <> 3

set k = m -' 4;
A10: ( F₅() . ((m -' 4) + 4) = F₇((m -' 4),(F₅() . (m -' 4)),(F₅() . ((m -' 4) + 1)),(F₅() . ((m -' 4) + 2)),(F₅() . ((m -' 4) + 3))) & F₆() . ((m -' 4) + 4) = F₇((m -' 4),(F₆() . (m -' 4)),(F₆() . ((m -' 4) + 1)),(F₆() . ((m -' 4) + 2)),(F₆() . ((m -' 4) + 3))) ) by A3, A6;
assume m <> 0 & ... & m <> 3 ; :: thesis: contradiction
then 3 < m ;
then 3 + 1 <= m by NAT_1:13;
then A11: m = (m -' 4) + 4 by XREAL_1:235;
then A12: F₅() . ((m -' 4) + 1) = F₆() . ((m -' 4) + 1) by A9, XREAL_1:6;
(m -' 4) + 3 < m by A11, XREAL_1:6;
then A13: F₅() . ((m -' 4) + 3) = F₆() . ((m -' 4) + 3) by A9;
(m -' 4) + 2 < m by A11, XREAL_1:6;
then A14: F₅() . ((m -' 4) + 2) = F₆() . ((m -' 4) + 2) by A9;
(m -' 4) + 0 < m by A11, XREAL_1:6;
hence contradiction by A8, A9, A11, A12, A14, A13, A10; :: thesis: verum

end;

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