How does Simplify resolve SimplifyCount ties?

Question

The expressions a (b + c) + b c and a b + (a + b) c are equivalent and equally complex as measured by SimplifyCount:

Reduce[a (b + c) + b c == a b + (a + b) c]
(* True *)

SimplifyCount[a (b + c) + b c] == SimplifyCount[a b + (a + b) c] == 9
(* True *)

And unlike the example that I used in How does `Simplify` resolve `LeafCount` ties?, the two expressions are not automatically simplified to the same expression before Simplify evaluates them:

a (b + c) + b c
(* b c + a (b + c) *)

a b + (a + b) c
(* a b + (a + b) c *)

a (b + c) + b c === a b + (a + b) c
(* False *)

Nevertheless, Simplify converts the latter expression into the former:

Simplify[a (b + c) + b c]
(* b c + a (b + c) *)

% === a (b + c) + b c
(* True *)

Simplify[a b + (a + b) c]
(* b c + a (b + c) *)

% === a b + (a + b) c
(* False *)

How does Simplify decide which expression is simpler, if they have the same SimplifyCount?

My guess is that it always breaks ties by going with whichever expression comes last in canonical order:

Sort[{a (b + c) + b c, a b + (a + b) c}]
(* {a b + (a + b) c, b c + a (b + c)} *)

Is this always the case? If so, then Simplify actually uses a total order on expressions, rather than the total preorder given by SimplifyCount.