User blog:Alemagno12/The problem with using X's in BEAF

Some people have tried to define their own interpretations of BEAF beyond ε0. The problem with those interpretations is that they use X arrays instead of the array of operator, which works fine for sub-legion arrays, but makes BEAF much weaker starting from {n,n/2}.

Using X arrays, {n,n/2} has growth rate ψ(Ωω), but using the array of operator, the growth rate of {n,m/2} depends on m: And eventually, we get that {n,m/2} (m = {n,n-1,2/2}) has growth rate ψ(Iω) = C(Ω) in Catching Function. Doing something that would add 1 to the ordinal when using X arrays changed the Ωω inside of psi function into an Iω when using the array of operator. Do you see now how replacing the array of operator makes BEAF a lot weaker?
 * {n,m/2} (m = n) has growth rate ψ(Ωω).
 * {n,m/2} (m = n+1) = n&{n,m/2} (m = n), so the m turns into an X, and {n,m/2} (m = n+1) has growth rate ψ(Ωω)+1.
 * {n,m/2} (m = n+2) = n&n&{n,m/2} (m = n), so the m turns into an X2, and {n,m/2} (m = n+1) has growth rate ψ(ΩΩ).
 * {n,m/2} (m = n+3) = n&n&n&{n,m/2} (m = n), so the m turns into an X3, and {n,m/2} (m = n+1) has growth rate ψ(ΩΩ 2 ).
 * {n,m/2} (m = 2n) = has growth rate ψ(ΩΩ ω ).
 * {n,m/2} (m = 3n) = has growth rate ψ(ΩΩ Ω ω ).
 * {n,m/2} (m = n2) = has growth rate ψ(ψI(0)).