above linear array arrays ({a,b,c,...} & base) then BEAF is really hard to define becuase of a new impasse.
Here I'll show how to continue, with a comparison to "beaf-extended cascading e"
{X,X(1)2} & 100 ~ E100{#,#|2}100
{X,X(1)2}^^X & 100 ~ E100{#,#|2}^^#100
{{X,X(1)2},X,1,2} & 100 ~ E100{{#,#|2},#,1,2}100
{{X,X|2},X|2} & 100 ~ E100{{#,#|2},#|2}100
{X,X+1|2} & 100 ~ E100{#,#|2}>#100
{X,X^X|2} & 100 ~ E100{#,#^#|2}100
{X,3,2|2} & 100 ~ E100{#,3,2|2}100
{X,X,2|2} & 100 ~ E100{#,#,2|2}100
{X,X,1,2|2} & 100 ~ E100{#,#,1,2|2}100
at this point a redefinition of the infinity barrier is in order...
{X,2,1,...,1|3} & 100 ~ E100{#,#|3}100
{X,2,1,...,1|X} & 100 ~ {X,2,1,...,1|1,2} ~ E100{#,#|#}100
{X,2,1,...,1|X^^X} & 100 ~ E100{#,#|#^^#}100
{X,2,1,...,1|{X,2,1,...,1|2}} & 100 ~ E100{#,#+1(1)2}3
{X,X,1,...,1|1,2} & 100 ~ {X,X+1(1)2} & 100 ~ E100{#,#+1(1)2}100
{X,X,1,...,1|1,1,2} & 100 ~ {X,X+2(1)2} & 100 ~ E100{#,#+2(1)2}100
{X,X,1,...,1|1,1,1,2} & 100 ~ {X,X+3(1)2} & 100 ~ E100{#,#+3(1)2}100
{X,2,1,...,1|1,...,1|2} & 100 ~ {X,2X(1)2} & 100 ~ E100{#,#+#(1)2}100
{X,2,1,...,1|1,...,1|1,...,1|2} & 100 ~ {X,3X(1)2} & 100 ~ E100{#,#+#+#(1)2}100
{X,2,1,...,1|1,...,1|1,...,1|... ... ...||2} & 100 ~ {X.X^2(1)2} & 100 ~ E100{#,##(1)2}100
{X,2,1,...,1|1,...,1|1,...,1|... ... ...||1,...,1|1,...,1|1,...,1|... ... ...||2} ~ {X,(X^2)*2(1)2} & 100
{X,2,1,...,1|1,...,1|1,...,1|... ... ...||1,...,1|1,...,1|1,...,1|... ... ...||1,...,1|1,...,1|1,...,1|... ... ...||... ... ... ... ... ... ... ... ...|||2} ~ {X,X^3(1)2} & 100
{X,X^4(1)2} & 100
{X,X^X(1)2} & 100
{X,X^^X(1)2} & 100
{X,3,2(1)2} & 100 = {X,{X,X(1)2}(1)2} & 100
{X,X,2(1)2} & 100
{X,{X,X,2(1)2}+1} & 100
{X,{X,{X,X,2(1)2}+1}+1} & 100
{X,X,2(1)2} | 2 & 100
{X,X,2(1)2} | 3 & 100
{X,X+1,2(1)2} & 100 = {X,X,2(1)2} | X & 100
{X,2X,2(1)2} & 100 = {X,X,2(1)2} | {X,X,2(1)2} & 100
{X,X^^X,2(1)2} & 100
{X,3,3(1)2} & 100
{X,X,3(1)2} & 100
...
For a while all you know from the past about linear arrays is enough until...
{X,X(1)X} & 100
{X,X,2(1)X+1} & 100
{X,2(1)X^^X} & 100
{X,3(1)1,2} & 100
{X,4(1)1,2} & 100
{X,X(1)1,2} & 100 ~ X_2 + 1 & X & 100