let p be Real; :: thesis: ( 0< p implies ex n being Nat st for m being Nat st n <= m holds |.((c . m)- g).|< p ) assume A6:
0< p
; :: thesis: ex n being Nat st for m being Nat st n <= m holds |.((c . m)- g).|< p then consider n1 being Nat such that A7:
for m being Nat st n1 <= m holds |.((a . m)- a1).|< p / 2
byA4; consider n2 being Nat such that A8:
for m being Nat st n2 <= m holds |.((b . m)- b1).|< p / 2
byA5, A6;
( n1 <= n1 + n2 & n2 <= n1 + n2 )
byNAT_1:11; then consider n being Nat such that A9:
n1 <= n
and A10:
n2 <= n
; take
n
; :: thesis: for m being Nat st n <= m holds |.((c . m)- g).|< p
for m being Nat st n <= m holds |.((c . m)- g).|< p