let n, m, k be Element of NAT ; ( PFArt (n,m) c= PFArt (k,m) implies n = k )
assume A1:
PFArt (n,m) c= PFArt (k,m)
; n = k
assume
n <> k
; contradiction
then
2 * n <> 2 * k
;
then A2:
[(2 * n),m] <> [(2 * k),m]
by XTUPLE_0:1;
[(2 * n),m] in PFArt (n,m)
by Def2;
then
ex m9 being odd Element of NAT st
( m9 <= 2 * k & [m9,m] = [(2 * n),m] )
by A1, A2, Def2;
hence
contradiction
by XTUPLE_0:1; verum