let m be Nat; for i being Integer ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>
let i be Integer; ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>
consider a being Element of m -tuples_on 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
;
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
A2:
len (2sComplement m,i) = m
by FINSEQ_1:def 18;
let j be
Nat;
( j in Seg m implies (2sComplement m,i) . j = a . j )
assume A3:
j in Seg m
;
(2sComplement m,i) . j = a . j
reconsider j =
j as
Nat ;
A4:
1
<= j
by A3, FINSEQ_1:3;
len a = m
by FINSEQ_1:def 18;
then A5:
j in dom a
by A3, FINSEQ_1:def 3;
A6:
j <= m
by A3, FINSEQ_1:3;
m <= m + 1
by NAT_1:11;
then
(
len (2sComplement (m + 1),i) = m + 1 &
j <= m + 1 )
by A6, FINSEQ_1:def 18, XXREAL_0:2;
then (2sComplement (m + 1),i) . j =
(2sComplement (m + 1),i) /. j
by A4, FINSEQ_4:24
.=
(2sComplement m,i) /. j
by A4, A6, Th32
.=
(2sComplement m,i) . j
by A2, A4, A6, FINSEQ_4:24
;
hence
(2sComplement m,i) . j = a . j
by A1, A5, FINSEQ_1:def 7;
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;
verum end; end; end;
hence
ex x being Element of BOOLEAN st 2sComplement (m + 1),i = (2sComplement m,i) ^ <*x*>
; verum