let m be Nat; :: thesis: for i being Integer ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>
let i be Integer; :: thesis: ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>
consider a being Tuple of m,BOOLEAN , b being Element of BOOLEAN such that
A1:
2sComplement (m + 1),i = a ^ <*b*>
by FINSEQ_2:137;
now per cases
( m > 0 or m = 0 )
;
suppose
m > 0
;
:: thesis: ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>then reconsider m =
m as non
empty Nat ;
for
j being
Nat st
j in Seg m holds
(2sComplement m,i) . j = a . j
proof
let j be
Nat;
:: thesis: ( j in Seg m implies (2sComplement m,i) . j = a . j )
assume A2:
j in Seg m
;
:: thesis: (2sComplement m,i) . j = a . j
reconsider j =
j as
Nat ;
A3:
(
len (2sComplement (m + 1),i) = m + 1 &
len (2sComplement m,i) = m &
len a = m )
by FINSEQ_1:def 18;
A4:
( 1
<= j &
j <= m )
by A2, FINSEQ_1:3;
m <= m + 1
by NAT_1:11;
then
j <= m + 1
by A4, XXREAL_0:2;
then A5:
(2sComplement (m + 1),i) . j =
(2sComplement (m + 1),i) /. j
by A3, A4, FINSEQ_4:24
.=
(2sComplement m,i) /. j
by A4, Th32
.=
(2sComplement m,i) . j
by A3, A4, FINSEQ_4:24
;
j in dom a
by A2, A3, FINSEQ_1:def 3;
hence
(2sComplement m,i) . j = a . j
by A1, A5, FINSEQ_1:def 7;
:: thesis: verum
end; then
a = 2sComplement m,
i
by FINSEQ_2:139;
hence
ex
x being
Element of
BOOLEAN st
2sComplement (m + 1),
i = (2sComplement m,i) ^ <*x*>
by A1;
:: thesis: verum end; end; end;
hence
ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>
; :: thesis: verum