Mixing
States that reach only bad states are bad. Predicate Logic
$begingroup$
Consider a system of states with a binary relation R. If R(x,y) holds, we say that state x reaches state y. Further, consider two unary predicates I and B where I(x) means that x is an initial state and B(x) means that x is a bad state.
Question: States that reach only bad states are bad.
Answer: ∀x(∀y(R(x,y) → B(y) → B(x)))
However, my question is in regard to "-> B(y)" shouldn't that be an and (∧),
∀x(∀y(R(x,y) ∧ B(y) → B(x)))
logic first-order-logic predicate-logic logic-translation
edited Jan 21 at 0:12
Bram28
63.2k44793
asked Jan 20 at 17:18
OctavianOctavian
83
$endgroup$
add a comment |
$begingroup$
Consider a system of states with a binary relation R. If R(x,y) holds, we say that state x reaches state y. Further, consider two unary predicates I and B where I(x) means that x is an initial state and B(x) means that x is a bad state.
Question: States that reach only bad states are bad.
Answer: ∀x(∀y(R(x,y) → B(y) → B(x)))
However, my question is in regard to "-> B(y)" shouldn't that be an and (∧),
∀x(∀y(R(x,y) ∧ B(y) → B(x)))
logic first-order-logic predicate-logic logic-translation
edited Jan 21 at 0:12
Bram28
63.2k44793
asked Jan 20 at 17:18
OctavianOctavian
83
$endgroup$
1
$begingroup$
The two are equivalent. See Exportation (logic).
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 17:23
$begingroup$
oh yeah, apologies the ∧ just read better in my head
$endgroup$
– Octavian
Jan 20 at 17:24
add a comment |
$begingroup$
Consider a system of states with a binary relation R. If R(x,y) holds, we say that state x reaches state y. Further, consider two unary predicates I and B where I(x) means that x is an initial state and B(x) means that x is a bad state.
Question: States that reach only bad states are bad.
Answer: ∀x(∀y(R(x,y) → B(y) → B(x)))
However, my question is in regard to "-> B(y)" shouldn't that be an and (∧),
∀x(∀y(R(x,y) ∧ B(y) → B(x)))
logic first-order-logic predicate-logic logic-translation
edited Jan 21 at 0:12
Bram28
63.2k44793
asked Jan 20 at 17:18
OctavianOctavian
83
$endgroup$
Consider a system of states with a binary relation R. If R(x,y) holds, we say that state x reaches state y. Further, consider two unary predicates I and B where I(x) means that x is an initial state and B(x) means that x is a bad state.
Question: States that reach only bad states are bad.
Answer: ∀x(∀y(R(x,y) → B(y) → B(x)))
However, my question is in regard to "-> B(y)" shouldn't that be an and (∧),
∀x(∀y(R(x,y) ∧ B(y) → B(x)))
logic first-order-logic predicate-logic logic-translation
logic first-order-logic predicate-logic logic-translation
edited Jan 21 at 0:12
Bram28
63.2k44793
asked Jan 20 at 17:18
OctavianOctavian
83
edited Jan 21 at 0:12
Bram28
63.2k44793
asked Jan 20 at 17:18
OctavianOctavian
83
edited Jan 21 at 0:12
Bram28
63.2k44793
edited Jan 21 at 0:12
Bram28
63.2k44793
edited Jan 21 at 0:12
Bram28
63.2k44793
63.2k44793
asked Jan 20 at 17:18
OctavianOctavian
83
asked Jan 20 at 17:18
OctavianOctavian
83
asked Jan 20 at 17:18
OctavianOctavian
83
83
1
$begingroup$
The two are equivalent. See Exportation (logic).
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 17:23
$begingroup$
oh yeah, apologies the ∧ just read better in my head
$endgroup$
– Octavian
Jan 20 at 17:24
add a comment |
1
$begingroup$
The two are equivalent. See Exportation (logic).
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 17:23
$begingroup$
oh yeah, apologies the ∧ just read better in my head
$endgroup$
– Octavian
Jan 20 at 17:24
1
1
$begingroup$
The two are equivalent. See Exportation (logic).
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 17:23
$begingroup$
The two are equivalent. See Exportation (logic).
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 17:23
$begingroup$
oh yeah, apologies the ∧ just read better in my head
$endgroup$
– Octavian
Jan 20 at 17:24
$begingroup$
oh yeah, apologies the ∧ just read better in my head
$endgroup$
– Octavian
Jan 20 at 17:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First, you need to add some parentheses, because $P rightarrow (Q rightarrow R)$ is not the same as $(P rightarrow Q) rightarrow R)$
Indeed, I see that you are hoping to use the Exportation equivalence rule, which states that
$$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$$
so that means that your intended symbolization must have been:
$$forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))) tag{*}$$
Unfortunately, $(*)$ is not the correct symbolization of the original sentence. $(*)$ translates back as: For any two states that you have, if the first reaches the second, then if the second is bad, then the first is bad. So, that would mean that as soon as the first state reaches any bad state, then the first state is bad. That is not the same as the original sentence, since that is only talking about states that only reach bad states. So, for example, take a state that reaches both good and bad states. Then by $(*)$, it must be a bad state, since it reaches at least one bad state. But the original English sentence has nothing to say about states that each both good and bad states.
OK, so how do you translate the original English sentence? Well, follow this form:
$$forall x (text{"x reaches only bad states"} rightarrow B(x))$$
OK, and the translation of "x reaches only bad states" is:
$$forall y (R(x,y) rightarrow B(y)$$
So, plug this in, and you get:
$$forall x (forall y (R(x,y) rightarrow B(y)) rightarrow B(x))$$
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
$endgroup$
add a comment |
$begingroup$
These two are equivalent due to the exportation rule within logic.
((P ∧ Q) -> R)) <=> (P -> (Q -> R))
answered Jan 20 at 17:39
OctavianOctavian
83
$endgroup$
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
First, you need to add some parentheses, because $P rightarrow (Q rightarrow R)$ is not the same as $(P rightarrow Q) rightarrow R)$
Indeed, I see that you are hoping to use the Exportation equivalence rule, which states that
$$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$$
so that means that your intended symbolization must have been:
$$forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))) tag{*}$$
Unfortunately, $(*)$ is not the correct symbolization of the original sentence. $(*)$ translates back as: For any two states that you have, if the first reaches the second, then if the second is bad, then the first is bad. So, that would mean that as soon as the first state reaches any bad state, then the first state is bad. That is not the same as the original sentence, since that is only talking about states that only reach bad states. So, for example, take a state that reaches both good and bad states. Then by $(*)$, it must be a bad state, since it reaches at least one bad state. But the original English sentence has nothing to say about states that each both good and bad states.
OK, so how do you translate the original English sentence? Well, follow this form:
$$forall x (text{"x reaches only bad states"} rightarrow B(x))$$
OK, and the translation of "x reaches only bad states" is:
$$forall y (R(x,y) rightarrow B(y)$$
So, plug this in, and you get:
$$forall x (forall y (R(x,y) rightarrow B(y)) rightarrow B(x))$$
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
$endgroup$
add a comment |
$begingroup$
First, you need to add some parentheses, because $P rightarrow (Q rightarrow R)$ is not the same as $(P rightarrow Q) rightarrow R)$
Indeed, I see that you are hoping to use the Exportation equivalence rule, which states that
$$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$$
so that means that your intended symbolization must have been:
$$forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))) tag{*}$$
Unfortunately, $(*)$ is not the correct symbolization of the original sentence. $(*)$ translates back as: For any two states that you have, if the first reaches the second, then if the second is bad, then the first is bad. So, that would mean that as soon as the first state reaches any bad state, then the first state is bad. That is not the same as the original sentence, since that is only talking about states that only reach bad states. So, for example, take a state that reaches both good and bad states. Then by $(*)$, it must be a bad state, since it reaches at least one bad state. But the original English sentence has nothing to say about states that each both good and bad states.
OK, so how do you translate the original English sentence? Well, follow this form:
$$forall x (text{"x reaches only bad states"} rightarrow B(x))$$
OK, and the translation of "x reaches only bad states" is:
$$forall y (R(x,y) rightarrow B(y)$$
So, plug this in, and you get:
$$forall x (forall y (R(x,y) rightarrow B(y)) rightarrow B(x))$$
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
$endgroup$
add a comment |
$begingroup$
First, you need to add some parentheses, because $P rightarrow (Q rightarrow R)$ is not the same as $(P rightarrow Q) rightarrow R)$
Indeed, I see that you are hoping to use the Exportation equivalence rule, which states that
$$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$$
so that means that your intended symbolization must have been:
$$forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))) tag{*}$$
Unfortunately, $(*)$ is not the correct symbolization of the original sentence. $(*)$ translates back as: For any two states that you have, if the first reaches the second, then if the second is bad, then the first is bad. So, that would mean that as soon as the first state reaches any bad state, then the first state is bad. That is not the same as the original sentence, since that is only talking about states that only reach bad states. So, for example, take a state that reaches both good and bad states. Then by $(*)$, it must be a bad state, since it reaches at least one bad state. But the original English sentence has nothing to say about states that each both good and bad states.
OK, so how do you translate the original English sentence? Well, follow this form:
$$forall x (text{"x reaches only bad states"} rightarrow B(x))$$
OK, and the translation of "x reaches only bad states" is:
$$forall y (R(x,y) rightarrow B(y)$$
So, plug this in, and you get:
$$forall x (forall y (R(x,y) rightarrow B(y)) rightarrow B(x))$$
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
$endgroup$
First, you need to add some parentheses, because $P rightarrow (Q rightarrow R)$ is not the same as $(P rightarrow Q) rightarrow R)$
Indeed, I see that you are hoping to use the Exportation equivalence rule, which states that
$$P rightarrow (Q rightarrow R) Leftrightarrow (P land Q) rightarrow R$$
so that means that your intended symbolization must have been:
$$forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))) tag{*}$$
Unfortunately, $(*)$ is not the correct symbolization of the original sentence. $(*)$ translates back as: For any two states that you have, if the first reaches the second, then if the second is bad, then the first is bad. So, that would mean that as soon as the first state reaches any bad state, then the first state is bad. That is not the same as the original sentence, since that is only talking about states that only reach bad states. So, for example, take a state that reaches both good and bad states. Then by $(*)$, it must be a bad state, since it reaches at least one bad state. But the original English sentence has nothing to say about states that each both good and bad states.
OK, so how do you translate the original English sentence? Well, follow this form:
$$forall x (text{"x reaches only bad states"} rightarrow B(x))$$
OK, and the translation of "x reaches only bad states" is:
$$forall y (R(x,y) rightarrow B(y)$$
So, plug this in, and you get:
$$forall x (forall y (R(x,y) rightarrow B(y)) rightarrow B(x))$$
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
edited Jan 21 at 17:10
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
answered Jan 21 at 0:06
Bram28Bram28
63.2k44793
63.2k44793
add a comment |
add a comment |
$begingroup$
These two are equivalent due to the exportation rule within logic.
((P ∧ Q) -> R)) <=> (P -> (Q -> R))
answered Jan 20 at 17:39
OctavianOctavian
83
$endgroup$
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
add a comment |
$begingroup$
These two are equivalent due to the exportation rule within logic.
((P ∧ Q) -> R)) <=> (P -> (Q -> R))
answered Jan 20 at 17:39
OctavianOctavian
83
$endgroup$
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
add a comment |
$begingroup$
These two are equivalent due to the exportation rule within logic.
((P ∧ Q) -> R)) <=> (P -> (Q -> R))
answered Jan 20 at 17:39
OctavianOctavian
83
$endgroup$
These two are equivalent due to the exportation rule within logic.
((P ∧ Q) -> R)) <=> (P -> (Q -> R))
answered Jan 20 at 17:39
OctavianOctavian
83
answered Jan 20 at 17:39
OctavianOctavian
83
answered Jan 20 at 17:39
OctavianOctavian
83
answered Jan 20 at 17:39
OctavianOctavian
83
83
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
add a comment |
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
Nope! Sorry, but your original statement is not of the right form to apply Exportation to. See my Answer.
$endgroup$
– Bram28
Jan 21 at 4:00
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@Bram28 The original statement (or rather a subformula of it) is of the right form to apply that equivalence assuming the OP meant the $to$s to associate to the right. What your answer implicitly points out is that OP's original formula does not correctly capture the intent of the natural language phrase.
$endgroup$
– Derek Elkins
Jan 21 at 5:40
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
$begingroup$
@DerekElkins Yes, quite right ... if the OP had meant $forall x (forall y (R(x,y) rightarrow (B(y) rightarrow B(x))))$, but in that case the symbolization contains two mistakes: the parentheses should go around the first $rightarrow$, and the scope of the $forall y$ should not extend to include the $B(x)$
$endgroup$
– Bram28
Jan 21 at 15:28
add a comment |
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1
$begingroup$
The two are equivalent. See Exportation (logic).
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 17:23
$begingroup$
oh yeah, apologies the ∧ just read better in my head
$endgroup$
– Octavian
Jan 20 at 17:24