theorem Th91:
  K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c= N
  implies K = 0 or exp(K,M) c= exp(L,N)
proof
  assume that
A1: K in L & M in N or K c= L & M in N or K in L & M c= N or K c= L & M c=
  N;
A2: K c= L by A1,CARD_1:3;
A3: M c= N by A1,CARD_1:3;
  now
    assume L <> {};
    then
A4: Funcs(N\M,L) <> {};
    0 c= card Funcs(N\M,L);
    then 0 in card Funcs(N\M,L) by A4,CARD_1:3;
    then
A5: nextcard Segm card 0 c= card Funcs(N\M,L) by CARD_1:def 3;
    0+1 = 1;
    then
A6: Segm 1 c= card Funcs(N\M,L) by A5,Lm8,NAT_1:42;
A7: M misses (N\M) by XBOOLE_1:79;
A8: N = M \/ (N\M) by A3,XBOOLE_1:45;
    Funcs(M,K) c= Funcs(M,L) by A2,FUNCT_5:56;
    then
A9: exp(K,M) c= card Funcs(M,L) by CARD_1:11;
A10: exp(L,N) = card [:Funcs(M,L),Funcs(N\M,L):] by A7,A8,FUNCT_5:62;
A11: card [:Funcs(M,L),Funcs(N\M,L):] = card [:card Funcs(M,L),card Funcs(N\
    M,L):] by Th6;
    card Funcs(M,L) *` card Funcs(N\M,L) = card [:card Funcs(M,L),card
    Funcs(N\M,L):];
    then 1 *` card Funcs(M,L) c= exp(L,N) by A6,A10,A11,Th89;
    then card Funcs(M,L) c= exp(L,N) by Th20;
    hence thesis by A9;
  end;
  hence thesis by A1;
end;
