consider k1 being Element of NAT such that A5:
x in A1 .((n1 + n2)+ k1)by A1, KURATO_2:8;
( x in A1 .((n1 + n2)+ k1) & not x in A2 .(n1 +(n2 + k1)) )
by A2, A5; then
( x in A1 .(n2 +(k1 + n1)) & not x in A2 .(n2 +(k1 + n1)) )
; hence
contradiction
by A4; :: thesis: verum