how to know all true in prolog

Question

Just started to learn Prolog, and cannot figure out how to get the result which involves "all true", for example:

preAction(4, 3).
preAction(4, 2).
preAction(2, 1).

action(4).
action(2).
action(1).

takeAction(X) :-
   action(X),
   preAction(X, Y),
   action(Y).

The expected result is 2, because action(2) and action(1) all true. But current result is 4 and 2. However the preAction of 4 is (3,2) and there is no action(3). This is not what I want.

If I tried to use negation, since 4's preAction is provable, still cannot figure out how to do it. Do you know how to do it?

`preAction` "of" 4 is not `(3,2)`. It's either `3` or `2`. And `action` is not `2` and `1`, but `2` **or** `1`. — Eugene Sh., Feb 27 '15 at 19:50
so, if I need to get the result of the actions that all preActions taken, how to do so ? — lMolly, Feb 27 '15 at 21:55

score 0 · Answer 1 · answered Feb 27 '15 at 22:28

You add another condition to your takeAction/1 predicate which uses negation:

takeAction(X) :-
    action(X),
    preAction(X, Y),
    action(Y),
    \+ (preAction(X, Z), \+ action(Z)).

The last line indicates that you want to exclude cases where X is a pre-action of Z, but Z is not an action (action(3) is false).

score 0 · Answer 2 · answered Feb 27 '15 at 23:22

0

You test that every preAction(X, Y) has an action(X) :

takeAction(X) :- 
   action(X), 
   forall(preAction(X, Y), action(Y)).

answered Feb 27 '15 at 23:22

joel76

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thank you! although we are required not to use any "high level function", still very helpful to correct my logical thinking. – lMolly Feb 28 '15 at 02:42
@lMolly It is a very strange requirement, not to use "high level function" in Prolog. Either way, keep in mind that `forall(preAction(X, Y), action(Y))` by definition behaves as `\+ (preAction(X, Y), \+ action(Y))`, which basically is the other solution you got. – Feb 28 '15 at 07:50

how to know all true in prolog

2 Answers2