Write in Logic: If professors are unhappy all students fail their exams
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I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:
∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]
However, the answer of my teacher is:
∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))
Can someone helps me?
logic first-order-logic
asked Dec 13 '18 at 9:22
GoldGold
61
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add a comment |
$begingroup$
I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:
∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]
However, the answer of my teacher is:
∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))
Can someone helps me?
logic first-order-logic
asked Dec 13 '18 at 9:22
GoldGold
61
$endgroup$
$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28
add a comment |
$begingroup$
I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:
∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]
However, the answer of my teacher is:
∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))
Can someone helps me?
logic first-order-logic
asked Dec 13 '18 at 9:22
GoldGold
61
$endgroup$
I have to write the following sentence "If professors are unhappy all students fail their exams" in logic and my answer is:
∀x [Prof(x) ∧ Unhappy(x)] ⇒ [∀y stud(y) ⇒ fail_exam(x,y)]
However, the answer of my teacher is:
∀x ∀y( prof(x) ∧ unhappy(x) ∧ stud(y) ) ⇒ fail exam(x, y))
Can someone helps me?
logic first-order-logic
logic first-order-logic
asked Dec 13 '18 at 9:22
GoldGold
61
asked Dec 13 '18 at 9:22
GoldGold
61
asked Dec 13 '18 at 9:22
GoldGold
61
asked Dec 13 '18 at 9:22
GoldGold
61
asked Dec 13 '18 at 9:22
GoldGold
61
61
$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28
add a comment |
$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28
$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28
$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:
$forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$
and
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$
Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:
$psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$
Here, the formula $psi$ cannot include any free variables $x$
Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$
Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:
$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$
Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:
$forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$
and
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$
Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:
$psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$
Here, the formula $psi$ cannot include any free variables $x$
Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$
Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:
$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$
Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:
$forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$
and
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$
Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:
$psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$
Here, the formula $psi$ cannot include any free variables $x$
Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$
Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:
$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$
Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
$endgroup$
add a comment |
$begingroup$
They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:
$forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$
and
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$
Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:
$psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$
Here, the formula $psi$ cannot include any free variables $x$
Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$
Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:
$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$
Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
$endgroup$
They are equivalent ... although to show that, I will first insist on adding a few parentheses so as to indicate the proper scope of the quantifiers, giving us:
$forall x ((Prof(x) land Unhappy(x)) rightarrow forall y (Stud(y) rightarrow FailExam(x,y)))$
and
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(y)) rightarrow FailExam(x,y))$
Now, to show these are equivalent, let us first note the following general 'Prenex Law', which is an equivalence that allows you to 'take out' quantifiers and broaden their scope to include ('move over') other parts of the formula:
$psi rightarrow forall x varphi(x) Leftrightarrow forall x (psi rightarrow varphi(x))$
Here, the formula $psi$ cannot include any free variables $x$
Well, we can apply this Prenex law to the second formula, and take out the $forall y$, since the antecedent of the conditional you are moving it over does not contain any free variables $y$. Thus, we get:
$forall x forall y ((Prof(x) land Unhappy(x) land Stud(x)) rightarrow FailExam(x,y))$
Ok, and now we can apply a second general equivalence principle that mAuro alluded to in the comments, called eXportation:
$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$
Applied to the previous formula, we thus obtain the first of your two formulas, thus showing that they are equivalent.
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
answered Dec 13 '18 at 13:12
Bram28Bram28
64.2k44793
64.2k44793
add a comment |
add a comment |
$begingroup$
You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
$endgroup$
add a comment |
$begingroup$
You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
$endgroup$
add a comment |
$begingroup$
You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
$endgroup$
You've been answered that they are equivalent. It's also worth knowing the transformation from $(Pland Q) rightarrow R$ to $Prightarrow Q rightarrow R$ is known by the name of "currying", and its inverse "uncurrying". It shows up not only in logic, but any domain that forms a Cartesian Closed Category.
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
answered Dec 13 '18 at 14:52
Jorge AdrianoJorge Adriano
59146
59146
add a comment |
add a comment |
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$begingroup$
The two are euivalent : $P to (Q to R)$ and $(P land Q) to R$ are equivalent.
$endgroup$
– Mauro ALLEGRANZA
Dec 13 '18 at 9:28