Mixing
Convert FOL clauses to PROLOG
2
I am very new to PROLOG so it might be a very trivial question, but I absolutely have no
idea how to solve it.
There are 4 sentences I need to formulate into PROLOG code:
All hounds howl at night.
Anyone who has any cats will not have any mice.
Light sleepers do not have anything which howls at night.
John has either a cat or a hound.
I convert sentences to well-formed formula in first-order predicate calculus like that
- ∀x (HOUND(x) → HOWL(x))
- ∀x ∀y (HAVE(x,y) ∧ CAT (y) → ¬∃z (HAVE(x,z) ∧ MOUSE (z)))
- ∀x (Light_Sleeper(x) → ¬∃y (HAVE (x,y) ∧ HOWL(y)))
- ∃x (HAVE (John,x) ∧ (CAT(x) ∨ HOUND(x)))
Now I don't know how to write them in Prolog.
Also, how can I make query on them?
prolog first-order-logic
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
asked Jan 1 at 9:31
NaghamNagham
111
add a comment |
2
I am very new to PROLOG so it might be a very trivial question, but I absolutely have no
idea how to solve it.
There are 4 sentences I need to formulate into PROLOG code:
All hounds howl at night.
Anyone who has any cats will not have any mice.
Light sleepers do not have anything which howls at night.
John has either a cat or a hound.
I convert sentences to well-formed formula in first-order predicate calculus like that
- ∀x (HOUND(x) → HOWL(x))
- ∀x ∀y (HAVE(x,y) ∧ CAT (y) → ¬∃z (HAVE(x,z) ∧ MOUSE (z)))
- ∀x (Light_Sleeper(x) → ¬∃y (HAVE (x,y) ∧ HOWL(y)))
- ∃x (HAVE (John,x) ∧ (CAT(x) ∨ HOUND(x)))
Now I don't know how to write them in Prolog.
Also, how can I make query on them?
prolog first-order-logic
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
asked Jan 1 at 9:31
NaghamNagham
111
The first clause is already a Horn clause, you can directly translate it.
– lambda.xy.x
Jan 2 at 9:41
Instead of using predicates to represent a cat / mouse / etc. you could use a constant. The second clause would become∀x ( Have(x,cat) → ¬ Have(x, mouse) ), but it is still not a Horn clause. Instead of defining a rule, it could be expressed as a query∀x ¬( Have(x,cat) ∧ Have(x, mouse) )but that needs negation. Since Prolog only implements negation as failure, the translation is only correct as long asHave(x,y)has only finitely many solutions.
– lambda.xy.x
Jan 2 at 9:51
The phrasing of the fourth clause suggests that John has a cat or a dog, but not both. If so, then you would need to make this explicit.
– lambda.xy.x
Jan 2 at 9:56
add a comment |
2
2
2
I am very new to PROLOG so it might be a very trivial question, but I absolutely have no
idea how to solve it.
There are 4 sentences I need to formulate into PROLOG code:
All hounds howl at night.
Anyone who has any cats will not have any mice.
Light sleepers do not have anything which howls at night.
John has either a cat or a hound.
I convert sentences to well-formed formula in first-order predicate calculus like that
- ∀x (HOUND(x) → HOWL(x))
- ∀x ∀y (HAVE(x,y) ∧ CAT (y) → ¬∃z (HAVE(x,z) ∧ MOUSE (z)))
- ∀x (Light_Sleeper(x) → ¬∃y (HAVE (x,y) ∧ HOWL(y)))
- ∃x (HAVE (John,x) ∧ (CAT(x) ∨ HOUND(x)))
Now I don't know how to write them in Prolog.
Also, how can I make query on them?
prolog first-order-logic
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
asked Jan 1 at 9:31
NaghamNagham
111
I am very new to PROLOG so it might be a very trivial question, but I absolutely have no
idea how to solve it.
There are 4 sentences I need to formulate into PROLOG code:
All hounds howl at night.
Anyone who has any cats will not have any mice.
Light sleepers do not have anything which howls at night.
John has either a cat or a hound.
I convert sentences to well-formed formula in first-order predicate calculus like that
- ∀x (HOUND(x) → HOWL(x))
- ∀x ∀y (HAVE(x,y) ∧ CAT (y) → ¬∃z (HAVE(x,z) ∧ MOUSE (z)))
- ∀x (Light_Sleeper(x) → ¬∃y (HAVE (x,y) ∧ HOWL(y)))
- ∃x (HAVE (John,x) ∧ (CAT(x) ∨ HOUND(x)))
Now I don't know how to write them in Prolog.
Also, how can I make query on them?
prolog first-order-logic
prolog first-order-logic
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
asked Jan 1 at 9:31
NaghamNagham
111
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
asked Jan 1 at 9:31
NaghamNagham
111
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
edited Jan 1 at 9:39
Brian Tompsett - 汤莱恩
4,2321338102
4,2321338102
asked Jan 1 at 9:31
NaghamNagham
111
asked Jan 1 at 9:31
NaghamNagham
111
asked Jan 1 at 9:31
NaghamNagham
111
111
The first clause is already a Horn clause, you can directly translate it.
– lambda.xy.x
Jan 2 at 9:41
Instead of using predicates to represent a cat / mouse / etc. you could use a constant. The second clause would become∀x ( Have(x,cat) → ¬ Have(x, mouse) ), but it is still not a Horn clause. Instead of defining a rule, it could be expressed as a query∀x ¬( Have(x,cat) ∧ Have(x, mouse) )but that needs negation. Since Prolog only implements negation as failure, the translation is only correct as long asHave(x,y)has only finitely many solutions.
– lambda.xy.x
Jan 2 at 9:51
The phrasing of the fourth clause suggests that John has a cat or a dog, but not both. If so, then you would need to make this explicit.
– lambda.xy.x
Jan 2 at 9:56
add a comment |
The first clause is already a Horn clause, you can directly translate it.
– lambda.xy.x
Jan 2 at 9:41
Instead of using predicates to represent a cat / mouse / etc. you could use a constant. The second clause would become∀x ( Have(x,cat) → ¬ Have(x, mouse) ), but it is still not a Horn clause. Instead of defining a rule, it could be expressed as a query∀x ¬( Have(x,cat) ∧ Have(x, mouse) )but that needs negation. Since Prolog only implements negation as failure, the translation is only correct as long asHave(x,y)has only finitely many solutions.
– lambda.xy.x
Jan 2 at 9:51
The phrasing of the fourth clause suggests that John has a cat or a dog, but not both. If so, then you would need to make this explicit.
– lambda.xy.x
Jan 2 at 9:56
The first clause is already a Horn clause, you can directly translate it.
– lambda.xy.x
Jan 2 at 9:41
The first clause is already a Horn clause, you can directly translate it.
– lambda.xy.x
Jan 2 at 9:41
Instead of using predicates to represent a cat / mouse / etc. you could use a constant. The second clause would become
∀x ( Have(x,cat) → ¬ Have(x, mouse) ), but it is still not a Horn clause. Instead of defining a rule, it could be expressed as a query ∀x ¬( Have(x,cat) ∧ Have(x, mouse) ) but that needs negation. Since Prolog only implements negation as failure, the translation is only correct as long as Have(x,y) has only finitely many solutions.– lambda.xy.x
Jan 2 at 9:51
Instead of using predicates to represent a cat / mouse / etc. you could use a constant. The second clause would become
∀x ( Have(x,cat) → ¬ Have(x, mouse) ), but it is still not a Horn clause. Instead of defining a rule, it could be expressed as a query ∀x ¬( Have(x,cat) ∧ Have(x, mouse) ) but that needs negation. Since Prolog only implements negation as failure, the translation is only correct as long as Have(x,y) has only finitely many solutions.– lambda.xy.x
Jan 2 at 9:51
The phrasing of the fourth clause suggests that John has a cat or a dog, but not both. If so, then you would need to make this explicit.
– lambda.xy.x
Jan 2 at 9:56
The phrasing of the fourth clause suggests that John has a cat or a dog, but not both. If so, then you would need to make this explicit.
– lambda.xy.x
Jan 2 at 9:56
add a comment |
1 Answer
1
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- All hounds howl at night.
hound(marshall).
hound(rubble).
howls_at_night(X) :-
hound(X).
%% ?- howls_at_night(everest).
%% false.
%% ?- howls_at_night(rubble).
%% true.
%% ?- howls_at_night(Name).
%% Name = marshall
%% Name = rubble.
- Anyone who has any cats will not have any mice.
cat(tom).
mice(jerry).
has(mammy, tom).
wont_have_mice(X) :- has(X, Y), cat(Y).
may_have_mice(X) :- has(X, Y) -> + cat(Y) ; true.
%% ?- may_have_mice(john).
%% true.
%% ?- wont_have_mice(john).
%% false.
%% ?- wont_have_mice(mammy).
%% true.
%% ?- may_have_mice(mammy).
%% false.
- Light sleepers do not have anything which howls at night.
has(ryder, marshall).
has(ryder, rubble).
could_be_a_lightsleeper(X) :- has(X, Y) -> + howls_at_night(Y) ; true.
is_not_a_lightsleeper(X) :- has(X, Y), howls_at_night(Y).
%% ?- could_be_a_lightsleeper(max).
%% true.
%% ?- could_be_a_lightsleeper(ryder).
%% false.
%% ?- could_be_a_lightsleeper(Name).
%% Name = mammy.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(mammy).
%% false.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(Name).
%% Name = ryder.
- John has either a cat or a hound.
has_cat_or_hound(X, Y) :- has(X, Y), (cat(Y) ; hound(Y)).
john_has(Y) :- cat(Y) ; hound(Y).
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
add a comment |
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
0
- All hounds howl at night.
hound(marshall).
hound(rubble).
howls_at_night(X) :-
hound(X).
%% ?- howls_at_night(everest).
%% false.
%% ?- howls_at_night(rubble).
%% true.
%% ?- howls_at_night(Name).
%% Name = marshall
%% Name = rubble.
- Anyone who has any cats will not have any mice.
cat(tom).
mice(jerry).
has(mammy, tom).
wont_have_mice(X) :- has(X, Y), cat(Y).
may_have_mice(X) :- has(X, Y) -> + cat(Y) ; true.
%% ?- may_have_mice(john).
%% true.
%% ?- wont_have_mice(john).
%% false.
%% ?- wont_have_mice(mammy).
%% true.
%% ?- may_have_mice(mammy).
%% false.
- Light sleepers do not have anything which howls at night.
has(ryder, marshall).
has(ryder, rubble).
could_be_a_lightsleeper(X) :- has(X, Y) -> + howls_at_night(Y) ; true.
is_not_a_lightsleeper(X) :- has(X, Y), howls_at_night(Y).
%% ?- could_be_a_lightsleeper(max).
%% true.
%% ?- could_be_a_lightsleeper(ryder).
%% false.
%% ?- could_be_a_lightsleeper(Name).
%% Name = mammy.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(mammy).
%% false.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(Name).
%% Name = ryder.
- John has either a cat or a hound.
has_cat_or_hound(X, Y) :- has(X, Y), (cat(Y) ; hound(Y)).
john_has(Y) :- cat(Y) ; hound(Y).
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
add a comment |
0
- All hounds howl at night.
hound(marshall).
hound(rubble).
howls_at_night(X) :-
hound(X).
%% ?- howls_at_night(everest).
%% false.
%% ?- howls_at_night(rubble).
%% true.
%% ?- howls_at_night(Name).
%% Name = marshall
%% Name = rubble.
- Anyone who has any cats will not have any mice.
cat(tom).
mice(jerry).
has(mammy, tom).
wont_have_mice(X) :- has(X, Y), cat(Y).
may_have_mice(X) :- has(X, Y) -> + cat(Y) ; true.
%% ?- may_have_mice(john).
%% true.
%% ?- wont_have_mice(john).
%% false.
%% ?- wont_have_mice(mammy).
%% true.
%% ?- may_have_mice(mammy).
%% false.
- Light sleepers do not have anything which howls at night.
has(ryder, marshall).
has(ryder, rubble).
could_be_a_lightsleeper(X) :- has(X, Y) -> + howls_at_night(Y) ; true.
is_not_a_lightsleeper(X) :- has(X, Y), howls_at_night(Y).
%% ?- could_be_a_lightsleeper(max).
%% true.
%% ?- could_be_a_lightsleeper(ryder).
%% false.
%% ?- could_be_a_lightsleeper(Name).
%% Name = mammy.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(mammy).
%% false.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(Name).
%% Name = ryder.
- John has either a cat or a hound.
has_cat_or_hound(X, Y) :- has(X, Y), (cat(Y) ; hound(Y)).
john_has(Y) :- cat(Y) ; hound(Y).
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
add a comment |
0
0
0
- All hounds howl at night.
hound(marshall).
hound(rubble).
howls_at_night(X) :-
hound(X).
%% ?- howls_at_night(everest).
%% false.
%% ?- howls_at_night(rubble).
%% true.
%% ?- howls_at_night(Name).
%% Name = marshall
%% Name = rubble.
- Anyone who has any cats will not have any mice.
cat(tom).
mice(jerry).
has(mammy, tom).
wont_have_mice(X) :- has(X, Y), cat(Y).
may_have_mice(X) :- has(X, Y) -> + cat(Y) ; true.
%% ?- may_have_mice(john).
%% true.
%% ?- wont_have_mice(john).
%% false.
%% ?- wont_have_mice(mammy).
%% true.
%% ?- may_have_mice(mammy).
%% false.
- Light sleepers do not have anything which howls at night.
has(ryder, marshall).
has(ryder, rubble).
could_be_a_lightsleeper(X) :- has(X, Y) -> + howls_at_night(Y) ; true.
is_not_a_lightsleeper(X) :- has(X, Y), howls_at_night(Y).
%% ?- could_be_a_lightsleeper(max).
%% true.
%% ?- could_be_a_lightsleeper(ryder).
%% false.
%% ?- could_be_a_lightsleeper(Name).
%% Name = mammy.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(mammy).
%% false.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(Name).
%% Name = ryder.
- John has either a cat or a hound.
has_cat_or_hound(X, Y) :- has(X, Y), (cat(Y) ; hound(Y)).
john_has(Y) :- cat(Y) ; hound(Y).
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
- All hounds howl at night.
hound(marshall).
hound(rubble).
howls_at_night(X) :-
hound(X).
%% ?- howls_at_night(everest).
%% false.
%% ?- howls_at_night(rubble).
%% true.
%% ?- howls_at_night(Name).
%% Name = marshall
%% Name = rubble.
- Anyone who has any cats will not have any mice.
cat(tom).
mice(jerry).
has(mammy, tom).
wont_have_mice(X) :- has(X, Y), cat(Y).
may_have_mice(X) :- has(X, Y) -> + cat(Y) ; true.
%% ?- may_have_mice(john).
%% true.
%% ?- wont_have_mice(john).
%% false.
%% ?- wont_have_mice(mammy).
%% true.
%% ?- may_have_mice(mammy).
%% false.
- Light sleepers do not have anything which howls at night.
has(ryder, marshall).
has(ryder, rubble).
could_be_a_lightsleeper(X) :- has(X, Y) -> + howls_at_night(Y) ; true.
is_not_a_lightsleeper(X) :- has(X, Y), howls_at_night(Y).
%% ?- could_be_a_lightsleeper(max).
%% true.
%% ?- could_be_a_lightsleeper(ryder).
%% false.
%% ?- could_be_a_lightsleeper(Name).
%% Name = mammy.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(mammy).
%% false.
%% ?- is_not_a_lightsleeper(max).
%% false.
%% ?- is_not_a_lightsleeper(Name).
%% Name = ryder.
- John has either a cat or a hound.
has_cat_or_hound(X, Y) :- has(X, Y), (cat(Y) ; hound(Y)).
john_has(Y) :- cat(Y) ; hound(Y).
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
answered Jan 2 at 14:40
Valera GrishinValera Grishin
34126
34126
add a comment |
add a comment |
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The first clause is already a Horn clause, you can directly translate it.
– lambda.xy.x
Jan 2 at 9:41
Instead of using predicates to represent a cat / mouse / etc. you could use a constant. The second clause would become
∀x ( Have(x,cat) → ¬ Have(x, mouse) ), but it is still not a Horn clause. Instead of defining a rule, it could be expressed as a query∀x ¬( Have(x,cat) ∧ Have(x, mouse) )but that needs negation. Since Prolog only implements negation as failure, the translation is only correct as long asHave(x,y)has only finitely many solutions.– lambda.xy.x
Jan 2 at 9:51
The phrasing of the fourth clause suggests that John has a cat or a dog, but not both. If so, then you would need to make this explicit.
– lambda.xy.x
Jan 2 at 9:56