Why the disjunctive ASP program cannot return both disjunct predicates instance in an answer?

Question

I am a newbie to ASP (Answer Set Programming).

I tried the following program in clingo online version (reasoning mode = enumerate all)

% instance
member(dave).

% encoding
edu(X);rich(X) :- member(X).

It returns:

Answer: 1
member(dave) rich(dave)
Answer: 2
member(dave) edu(dave)
SATISFIABLE

But why there is no "Answer: 3" ? Something like:

Answer: 3
member(dave) edu(dave) rich(dave)

This answer also satisfies the rule edu(X);rich(X) :- member(X)..

From the perspective of logic,

edu(X);rich(X) :- member(X).

means

member(X) -> edu(X) ∨ rich(X)

So there ought to be another answer that both edu(X) and rich(X) are true.

BTW, if I add the two facts about charlie manually:

% instance
member(charlie).
rich(charlie).
edu(charlie).

% encoding
edu(X);rich(X) :- member(X).

Then it returns the correct answer as expected:

Answer: 1
member(charlie) edu(charlie) rich(charlie)
SATISFIABLE

Very thanks!

---

Some background information:

I have this question, because I am comparing traditional Logic Programming (e.g Prolog) vs Answer Set Programming (e.g. clingo), although they use different theoretical framework.

The disjunctive programs of the form in clingo:

a_1 ∨ a_2 ... ∨ a_n <- body

can not be represented in Prolog out of box.

A related question, see Using And clause in HEAD of a prolog statement

Answer 1

It seems that we can use the form Choice Rule to represent the logical disjunctive semantics in clingo.

Example 1:

% instance
member(dave).

% encoding
1 {edu(X);rich(X)} :- member(X).

returns

Answer: 1
member(dave) edu(dave)
Answer: 2
member(dave) rich(dave)
Answer: 3
member(dave) rich(dave) edu(dave)
SATISFIABLE

Example 2:

% instance
member(dave).
member(charlie).
rich(charlie).
edu(charlie).

% encoding
1 {edu(X);rich(X)} :- member(X).

returns

Answer: 1
edu(charlie) rich(charlie) member(dave) member(charlie) edu(dave)
Answer: 2
edu(charlie) rich(charlie) member(dave) member(charlie) rich(dave)
Answer: 3
edu(charlie) rich(charlie) member(dave) member(charlie) rich(dave) edu(dave)
SATISFIABLE

It works as expected, although I have no idea whether is it correct semantically.