let m, k, l be Nat; ( k <> l & 1 <= k & k <= m & 1 <= l & l <= m implies ((m !) * k) + 1,((m !) * l) + 1 are_coprime )
assume S1:
( k <> l & 1 <= k & k <= m & 1 <= l & l <= m )
; ((m !) * k) + 1,((m !) * l) + 1 are_coprime
assume
not ((m !) * k) + 1,((m !) * l) + 1 are_coprime
; contradiction
then consider d being non zero Nat such that
A1:
( d <> 1 & d divides ((m !) * k) + 1 & d divides ((m !) * l) + 1 )
by MOEBIUS2:5;
per cases
( k < l or k > l )
by S1, XXREAL_0:1;
suppose
k < l
;
contradictionthen b1:
l - k > k - k
by XREAL_1:14;
a1:
d divides l * (((m !) * k) + 1)
by A1, NAT_D:9;
d divides k * (((m !) * l) + 1)
by A1, NAT_D:9;
then A2:
d divides (l * (((m !) * k) + 1)) - (k * (((m !) * l) + 1))
by a1, PREPOWER:94;
A3:
l - k <= m - k
by S1, XREAL_1:13;
m - k < m
by S1, XREAL_1:44;
then A5:
l - k < m
by A3, XXREAL_0:2;
d <= l - k
by b1, A2, NAT_D:7;
then
d <= m
by A5, XXREAL_0:2;
then
d divides (m !) * k
by NAT_D:9, NEWTON:41;
hence
contradiction
by A1, WSIERP_1:15, NAT_D:10;
verum end; suppose
k > l
;
contradictionthen b1:
k - l > l - l
by XREAL_1:14;
a1:
d divides l * (((m !) * k) + 1)
by A1, NAT_D:9;
d divides k * (((m !) * l) + 1)
by A1, NAT_D:9;
then A2:
d divides (k * (((m !) * l) + 1)) - (l * (((m !) * k) + 1))
by a1, PREPOWER:94;
A3:
k - l <= m - l
by S1, XREAL_1:13;
m - l < m
by S1, XREAL_1:44;
then A5:
k - l < m
by A3, XXREAL_0:2;
d <= k - l
by b1, A2, NAT_D:7;
then
d <= m
by A5, XXREAL_0:2;
then
d divides (m !) * k
by NAT_D:9, NEWTON:41;
hence
contradiction
by A1, WSIERP_1:15, NAT_D:10;
verum end; end;