let E be set ; for A being Subset of (E ^omega)
for m, n being Nat holds
( A |^ (m,n) = {} iff ( m > n or ( m > 0 & A = {} ) ) )
let A be Subset of (E ^omega); for m, n being Nat holds
( A |^ (m,n) = {} iff ( m > n or ( m > 0 & A = {} ) ) )
let m, n be Nat; ( A |^ (m,n) = {} iff ( m > n or ( m > 0 & A = {} ) ) )
thus
( not A |^ (m,n) = {} or m > n or ( m > 0 & A = {} ) )
( ( m > n or ( m > 0 & A = {} ) ) implies A |^ (m,n) = {} )proof
assume that A3:
A |^ (
m,
n)
= {}
and A4:
(
m <= n & (
m <= 0 or
A <> {} ) )
;
contradiction
now ( m <= n implies not m = 0 )end;
hence
contradiction
by A4, A5;
verum
end;
hence
( ( m > n or ( m > 0 & A = {} ) ) implies A |^ (m,n) = {} )
by A1; verum