f
(
n
0
)
=
m
0
{\displaystyle f(n_{0})=m_{0}}
f
(
n
)
=
m
1
{\displaystyle f(n)=m_{1}}
for all
n
{\displaystyle n}
's which are different from
n
0
{\displaystyle n_{0}}
.
The totality of different coverings of N with M forms a definite aggregate with the elements
f
(
N
)
{\displaystyle f(N)}
;
we call it the "covering-aggregate (Belegungsmenge ) of
N
{\displaystyle N}
with
M
{\displaystyle M}
" and denote it by
(
N
|
M
)
{\displaystyle (N|M)}
. Thus:
(2)
(
N
|
M
)
=
f
(
N
)
{\displaystyle (N|M)={f(N)}}
.
If
M
∼
M
′
{\displaystyle M\sim M'}
and
N
∼
N
′
{\displaystyle N\sim N'}
, we easily find that
(3)
(
N
|
M
)
∼
(
N
′
|
M
′
)
{\displaystyle (N|M)\sim (N'|M')}
.
Thus the cardinal number of
(
N
|
M
)
{\displaystyle (N|M)}
depends only on the cardinal numbers
M
¯
¯
=
a
{\displaystyle {\overline {\overline {M}}}={\mathfrak {a}}}
and
N
¯
¯
=
b
{\displaystyle {\overline {\overline {N}}}={\mathfrak {b}}}
; it serves us for the definition of
a
b
{\displaystyle {\mathfrak {a}}^{\mathfrak {b}}}
:
(4)
a
b
=
(
N
|
M
¯
¯
)
{\displaystyle {\mathfrak {a}}^{\mathfrak {b}}=({\overline {\overline {N|M}}})}
.
For any three aggregates,
M
,
N
,
P
{\displaystyle M,N,P}
, we easily prove the theorems:
(5)
(
(
N
|
M
)
.
(
P
|
M
)
)
∼
(
(
N
.
P
)
|
M
)
{\displaystyle ((N|M).(P|M))\sim ((N.P)|M)}
,
(6)
(
(
P
|
M
)
.
(
P
|
N
)
)
∼
(
P
|
(
M
.
N
)
)
{\displaystyle ((P|M).(P|N))\sim (P|(M.N))}
,
(7)
(
P
|
(
N
|
M
)
)
∼
(
(
P
.
N
)
|
M
)
{\displaystyle (P|(N|M))\sim ((P.N)|M)}
,
from which, if we put
P
¯
¯
=
c
{\displaystyle {\overline {\overline {P}}}={\mathfrak {c}}}
, we have, by (4) and by
paying attention to § 3, the theorems for any three cardinal numbers,
a
{\displaystyle {\mathfrak {a}}}
,
b
{\displaystyle {\mathfrak {b}}}
, and
c
{\displaystyle {\mathfrak {c}}}
:
(8)
a
b
.
a
c
=
a
b
+
c
{\displaystyle {\mathfrak {a}}^{\mathfrak {b}}.{\mathfrak {a}}^{\mathfrak {c}}={\mathfrak {a}}^{\mathfrak {b+c}}}
,
(9)
a
c
.
b
c
=
(
a
.
b
)
c
{\displaystyle {\mathfrak {a}}^{\mathfrak {c}}.{\mathfrak {b}}^{\mathfrak {c}}=({\mathfrak {a}}.{\mathfrak {b}})^{\mathfrak {c}}}
,
(10)
(
a
b
)
c
=
a
b
.
c
{\displaystyle ({\mathfrak {a}}^{\mathfrak {b}})^{\mathfrak {c}}={\mathfrak {a}}^{\mathfrak {b.c}}}
.