let n be Nat; :: thesis: ( n > 0 implies for a being Element of Segm n holds
( ( a = 0 implies (compint n) . a = 0 ) & ( (compint n) . a = 0 implies a = 0 ) & ( a <> 0 implies (compint n) . a = n - a ) & ( (compint n) . a = n - a implies a <> 0 ) ) )
assume A1: n > 0 ; :: thesis: for a being Element of Segm n holds
( ( a = 0 implies (compint n) . a = 0 ) & ( (compint n) . a = 0 implies a = 0 ) & ( a <> 0 implies (compint n) . a = n - a ) & ( (compint n) . a = n - a implies a <> 0 ) )
let a be Element of Segm n; :: thesis: ( ( a = 0 implies (compint n) . a = 0 ) & ( (compint n) . a = 0 implies a = 0 ) & ( a <> 0 implies (compint n) . a = n - a ) & ( (compint n) . a = n - a implies a <> 0 ) )
reconsider n = n as non zero Nat by A1;
reconsider a = a as Element of NAT by ORDINAL1:def 12;
A2: a < n by NAT_1:44;
then a - a < n - a by XREAL_1:9;
then reconsider b = n - a as Element of NAT by INT_1:3;
consider c being Element of NAT such that
A3: c = b mod n ;
A4: ( (compint n) . a = 0 implies a = 0 )

proof

a - a < n - a by A2, XREAL_1:9;
then reconsider a9 = n - a as Element of NAT by INT_1:3;
assume A5: (compint n) . a = 0 ; :: thesis: a = 0
n <= n + a by NAT_1:11;
then A6: n - a <= (n + a) - a by XREAL_1:9;
consider t being Nat such that
A7: a9 = (n * t) + (a9 mod n) and
a9 mod n < n by NAT_D:def 2;
assume a <> 0 ; :: thesis: contradiction
then n - a <> n ;
then A8: n - a < n by A6, XXREAL_0:1;
t = 0

proof

assume t <> 0 ; :: thesis: contradiction
then 1 + 0 <= t by INT_1:7;
then A9: 1 * n <= t * n by XREAL_1:64;
t * n <= (t * n) + (a9 mod n) by NAT_1:11;
hence contradiction by A8, A7, A9, XXREAL_0:2; :: thesis: verum

end;

then a9 = 0 by A5, A7, Def11;
hence contradiction by NAT_1:44; :: thesis: verum

end;

consider t being Nat such that
A10: ( ( b = (n * t) + c & c < n ) or ( c = 0 & n = 0 ) ) by A3, NAT_D:def 2;
A11: n - a <= n

proof

assume n - a > n ; :: thesis: contradiction
then (n - a) + a > n + a by XREAL_1:6;
hence contradiction by NAT_1:11; :: thesis: verum

end;

A12: now :: thesis: ( a = 0 implies (compint n) . a = 0 )

assume A13: a = 0 ; :: thesis: (compint n) . a = 0
A14: t = 1

proof

now :: thesis: ( ( t = 0 & t = 1 ) or ( t <> 0 & t = 1 ) )

per cases ( t = 0 or t <> 0 ) ;

case t = 0 ; :: thesis: t = 1

hence t = 1 by A10, A13; :: thesis: verum

end;

case A15: t <> 0 ; :: thesis: t = 1

t < 2

proof

assume t >= 2 ; :: thesis: contradiction
then A16: n * t >= n * 2 by XREAL_1:64;
A17: n <= (n * 1) + (n * 1) by NAT_1:11;
(n * t) + c >= n * t by NAT_1:11;
then n - a >= n * 2 by A10, A16, XXREAL_0:2;
then n * 1 = 2 * n by A13, A17, XXREAL_0:1;
hence contradiction by A1; :: thesis: verum

end;

then t < 1 + 1 ;
then A18: t <= 1 by NAT_1:13;
1 + 0 <= t by A15, INT_1:7;
hence t = 1 by A18, XXREAL_0:1; :: thesis: verum

end;

end;

end;

hence t = 1 ; :: thesis: verum

end;

c = 0

proof

assume c <> 0 ; :: thesis: contradiction
then n + c > n + 0 by XREAL_1:6;
hence contradiction by A10, A11, A14; :: thesis: verum

end;

hence (compint n) . a = 0 by A3, Def11; :: thesis: verum

end;

now :: thesis: ( a <> 0 implies (compint n) . a = n - a )

assume A19: a <> 0 ; :: thesis: (compint n) . a = n - a
A20: n - a < n

proof

assume n - a >= n ; :: thesis: contradiction
then n - a = n by A11, XXREAL_0:1;
hence contradiction by A19; :: thesis: verum

end;

t = 0

proof

assume t <> 0 ; :: thesis: contradiction
then 1 + 0 <= t by INT_1:7;
then A21: 1 * n <= t * n by XREAL_1:64;
n * t <= (n * t) + c by NAT_1:11;
hence contradiction by A10, A20, A21, XXREAL_0:2; :: thesis: verum

end;

hence (compint n) . a = n - a by A3, A10, Def11; :: thesis: verum

end;

hence ( ( a = 0 implies (compint n) . a = 0 ) & ( (compint n) . a = 0 implies a = 0 ) & ( a <> 0 implies (compint n) . a = n - a ) & ( (compint n) . a = n - a implies a <> 0 ) ) by A12, A4; :: thesis: verum