let m, n be Nat; ( Seg n, Seg m are_equipotent implies n = m )
defpred S2[ Nat] means ex n being Nat st
( Seg n, Seg $1 are_equipotent & n <> $1 );
assume
( Seg n, Seg m are_equipotent & n <> m )
; contradiction
then A1:
ex m being Nat st S2[m]
;
consider m being Nat such that
A2:
S2[m]
and
A3:
for k being Nat st S2[k] holds
m <= k
from NAT_1:sch 5(A1);
consider n being Nat such that
A4:
Seg n, Seg m are_equipotent
and
A5:
n <> m
by A2;
then consider m1 being Nat such that
A8:
m = m1 + 1
by NAT_1:6;
then consider n1 being Nat such that
A10:
n = n1 + 1
by NAT_1:6;
A11:
n in Seg n
by A9, Th3;
A12:
m in Seg m
by A7, Th3;
A13:
not n1 + 1 <= n1
by NAT_1:13;
A14:
not m1 + 1 <= m1
by NAT_1:13;
A15:
not n in Seg n1
by A10, A13, Th1;
A16:
not m in Seg m1
by A8, A14, Th1;
A17:
(Seg n1) /\ {n} c= {}
A18:
(Seg m1) /\ {m} c= {}
A19:
Seg n = (Seg n1) \/ {n}
by A10, Th9;
A20:
Seg m = (Seg m1) \/ {m}
by A8, Th9;
A23:
( (Seg n1) \ {n} = ((Seg n1) \/ {n}) \ {n} & Seg n1 misses {n} )
by A17, XBOOLE_1:40;
A24:
( (Seg m1) \ {m} = ((Seg m1) \/ {m}) \ {m} & Seg m1 misses {m} )
by A18, XBOOLE_1:40;
A25:
(Seg n) \ {n} = Seg n1
by A19, A23, XBOOLE_1:83;
(Seg m) \ {m} = Seg m1
by A20, A24, XBOOLE_1:83;
hence
contradiction
by A3, A4, A5, A8, A10, A11, A12, A14, A25, CARD_1:34; verum