Then it is equivalent to the thing I call B2 in edit 2 in the post (Assuming A1-A3).
In this case, your modified B2 is my B2, and your B3 is my A4, which follows from A5 assuming A1-A3 and B2, so your suspicion that these imply C4 is stronger than my Q6, which is false, as I argue here.
However, without A5, it is actually much easier to see that this doesn’t work. The counterexample here satisfies my A1-A3, your weaker version of B2, your B3, and violates C4.
I propose the axioms A1-A3 together with
B2. If A≺B then for any p∈(0,1) we have A≺pA+(1−p)B≺B
B3. If A⪯B and A⪯C, then for any p∈[0,1] we have A⪯pB+(1−p)C
I suspect that these imply C4.
Your B2 is going to rule out a bunch of concave functions. I was hoping to only use axioms consistent with all (continuous) concave functions.
Oops. What if instead of “for any p” we go with “there exists p”?
Then it is equivalent to the thing I call B2 in edit 2 in the post (Assuming A1-A3).
In this case, your modified B2 is my B2, and your B3 is my A4, which follows from A5 assuming A1-A3 and B2, so your suspicion that these imply C4 is stronger than my Q6, which is false, as I argue here.
However, without A5, it is actually much easier to see that this doesn’t work. The counterexample here satisfies my A1-A3, your weaker version of B2, your B3, and violates C4.
Your B3 is equivalent to A4 (assuming A1-3).