let L, K, M, N be Cardinal; ( K in L & M in N implies ( K *` M in L *` N & M *` K in L *` N ) )
A1:
for K, L, M, N being Cardinal st K in L & M in N & L c= N holds
K *` M in L *` N
proof
let K,
L,
M,
N be
Cardinal;
( K in L & M in N & L c= N implies K *` M in L *` N )
assume that A2:
K in L
and A3:
M in N
and A4:
L c= N
;
K *` M in L *` N
A5:
now ( N is finite implies K *` M in L *` N )assume A6:
N is
finite
;
K *` M in L *` Nthen reconsider N =
N as
finite Cardinal ;
reconsider L =
L,
M =
M,
K =
K as
finite Cardinal by A2, A3, A4, A6, CARD_3:92;
A7:
card (Segm N) = N
;
card (Segm M) = M
;
then
card M < card N
by A3, A7, NAT_1:41;
then A8:
(card K) * (card M) <= (card K) * (card N)
by XREAL_1:64;
A9:
card (Segm L) = L
;
A10:
L *` N = card (Segm ((card L) * (card N)))
by CARD_2:39;
card (Segm K) = K
;
then
card K < card L
by A2, A9, NAT_1:41;
then
(card K) * (card N) < (card L) * (card N)
by A3, XREAL_1:68;
then A11:
(card K) * (card M) < (card L) * (card N)
by A8, XXREAL_0:2;
K *` M = card (Segm ((card K) * (card M)))
by CARD_2:39;
hence
K *` M in L *` N
by A10, A11, NAT_1:41;
verum end;
A12:
0 in L
by A2, ORDINAL3:8;
hence
K *` M in L *` N
by A5;
verum
end;
( L c= N or N c= L )
;
hence
( K in L & M in N implies ( K *` M in L *` N & M *` K in L *` N ) )
by A1; verum