let seq1, seq2, seq3 be Complex_Sequence; :: thesis: ( seq1 = seq2 - seq3 iff for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n) )
A1: ( ( for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n) ) implies seq1 = seq2 - seq3 )
proof
assume A2: for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n) ; :: thesis: seq1 = seq2 - seq3
for x being object st x in NAT holds
seq1 . x = (seq2 - seq3) . x
proof
let x be object ; :: thesis: ( x in NAT implies seq1 . x = (seq2 - seq3) . x )
assume x in NAT ; :: thesis: seq1 . x = (seq2 - seq3) . x
then reconsider x = x as Element of NAT ;
seq1 . x = (seq2 . x) - (seq3 . x) by A2
.= (seq2 . x) + (- (seq3 . x))
.= (seq2 . x) + ((- seq3) . x) by VALUED_1:8
.= (seq2 + (- seq3)) . x by VALUED_1:1 ;
hence seq1 . x = (seq2 - seq3) . x ; :: thesis: verum
end;
hence seq1 = seq2 - seq3 by FUNCT_2:12; :: thesis: verum
end;
( seq1 = seq2 - seq3 implies for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n) )
proof
assume A3: seq1 = seq2 - seq3 ; :: thesis: for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n)
now :: thesis: for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n)
let n be Nat; :: thesis: seq1 . n = (seq2 . n) - (seq3 . n)
A4: n in NAT by ORDINAL1:def 12;
seq1 . n = (seq2 . n) + ((- seq3) . n) by
.= (seq2 . n) + (- (seq3 . n)) by VALUED_1:8 ;
hence seq1 . n = (seq2 . n) - (seq3 . n) ; :: thesis: verum
end;
hence for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n) ; :: thesis: verum
end;
hence ( seq1 = seq2 - seq3 iff for n being Nat holds seq1 . n = (seq2 . n) - (seq3 . n) ) by A1; :: thesis: verum