let k be Nat; for i1, i2 being Integer st 2 <= k & i1 in k -SD & i2 in k -SD_Sub holds
SDSub_Add_Data ((i1 + i2),k) in k -SD_Sub_S
let i1, i2 be Integer; ( 2 <= k & i1 in k -SD & i2 in k -SD_Sub implies SDSub_Add_Data ((i1 + i2),k) in k -SD_Sub_S )
assume that
A1:
2 <= k
and
A2:
( i1 in k -SD & i2 in k -SD_Sub )
; SDSub_Add_Data ((i1 + i2),k) in k -SD_Sub_S
set z = i1 + i2;
( i1 <= (Radix k) - 1 & i2 <= Radix (k -' 1) )
by A2, Th1, RADIX_1:13;
then A3:
i1 + i2 <= ((Radix k) - 1) + (Radix (k -' 1))
by XREAL_1:7;
( (- (Radix k)) + 1 <= i1 & (- (Radix (k -' 1))) - 1 <= i2 )
by A2, Th1, RADIX_1:13;
then A4:
((- (Radix k)) + 1) + ((- (Radix (k -' 1))) - 1) <= i1 + i2
by XREAL_1:7;
A5:
( SDSub_Add_Data ((i1 + i2),k) >= - (Radix (k -' 1)) & SDSub_Add_Data ((i1 + i2),k) <= (Radix (k -' 1)) - 1 )
SDSub_Add_Data ((i1 + i2),k) is Element of INT
by INT_1:def 2;
hence
SDSub_Add_Data ((i1 + i2),k) in k -SD_Sub_S
by A5; verum