(a)

x⟡y does not appear to be commutative in general because sandwiching preserves the order of x and y. The unital property is a special case of x⟡y. Just because x⟡1=1⟡x it does not follow that replacing 1 by y justifies commutativity.

(b)

Since 1∊S and x=(x⟡1), x⟡(y⟡z)=(x⟡1)⟡(y⟡z)=(x⟡y)⟡(1⟡z)=(x⟡y)⟡z. Associativity is implied.

(c)

No. We can say that x⦁(y⦁z)=(x⦁y)⦁(x⦁z) but we can’t evaluate (x⦁y)⦁z from the self-distributive rule alone (without sandwiching). The unital rule doesn’t provide any aid in this case, if we replace x, y or z with 1.

(d)

We can add the sandwiching rule to (c) and prove associativity for the ⦁ operation, following the reasoning in (b), but we arrive at x⦁(y⦁z)=(x⦁1)⦁(y⦁z)=(x⦁y)⦁(1⦁z)=(x⦁y)⦁z, and x⦁(y⦁z)=(x⦁y)⦁(x⦁z). Therefore we get:

(x⦁y)⦁z=(x⦁y)⦁(x⦁z) rather than (x⦁z)⦁(y⦁z) or (z⦁x)⦁(z⦁y) which we might have expected.

The self-distributive property seems to be unnatural (while unital, sandwiching, commutative and associative properties are natural. The unital property is more naturally and generally applicable if extended to include any y∊S instead of restricting to 1; but x⦁y is only x or y if y or x is 1. If we inspect the standard + and × operations, for example, we know the unital, associative and commutative rules apply. Sandwiching also applies, but the self-distributive rule does not apply:

x+(y+z)≠(x+y)+(x+z), x×(y×z)≠(x×y)×(x×z).

The rules are also extendable to numbers generally, not just the natural numbers.

Let ⋆ be defined as x⋆y=x+y+1, then x⋆y=y⋆x (commutative); x⋆(y⋆z)=x⋆(y+z+1)=x+y+z+2, and (x⋆y)⋆z=(x+y+1)⋆z=x+y+z+2 (associative). x⋆1=1⋆x=x+2, not x, so the unital rule doesn’t apply. Neither does the self-distributive rule.

If we substitute 1 in place of x, y or z these rules for × but not +, because x+1≠x.