[perl #132713] Order of `is assoc` and precedence traits matter, but should it?

[brian d foy]

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I originally asked this on StackOverflow:

    https://stackoverflow.com/q/48219788/2766176

This fails to be right associative:

    sub infix:<↑> ( Int:D \n, Int:D \m  --> Int:D )
        is assoc<right>
        is equiv(&infix:<**>)
        { n ** m }

    put "2**2**2**2 = ",      2**2**2**2;
    put "2↑2↑2↑2 = ",         2↑2↑2↑2;
    put "2↑ (2↑ (2↑2) ) = ",  2↑ (2↑ (2↑2) );

Reversing the traits fixes that:

    sub infix:<↑> ( Int:D \n, Int:D \m  --> Int:D )
        is equiv(&infix:<**>)
        is assoc<right>
        { n ** m }

I didn't see anything in the docs about the ordering of traits but
I wouldn't expect order to matter. Neither did I see anything saying I
couldn't combine traits.

As far as I can tell the precedence still works when it's specified first:

    sub infix:<↑> ( Int:D \n, Int:D \m  --> Int:D )
        is equiv(&infix:<**>)
        is assoc<right>
        { n ** m }

    put "2↑3**4↑2 = ",         2↑2**2↑2;
    put "2↑(3**(4↑2)) = ",         2↑2**2↑2;


    put "2↑3*4↑2 = ",         2↑2*2↑2;
    put "(2↑2)*(2↑2) = ",        (2↑2)*(2↑2);


    put "2↑3+4↑2 = ",         2↑2+2↑2;
    put "(2↑2)+(2↑2) = ",         (2↑2)+(2↑2);

How is this supposed to work? Should either way work? I can easily
imagine situations where I want to stack many traits:

    sub infix:<↑>  ( Int:D \n, Int:D \m  --> Int:D )
        is equiv(&infix:<**>)
        is assoc<right>
        is pure
        is export
        { ... }