classical logic - rules for quantifiers The Next CEO of Stack OverflowCan second order logic express each (computable) infinitary logic sentence?Logic coursework assistanceSubstitution not clear in logicStuck on a quantifier logic problemAre proofs for many-sorted first order logic shorter than single sorted first order logic?Prove each equivalence by using the rules for semantic equivalenceClassical logic without negation and falsehoodCurious tableaux rules for a curious logicpredicate logic proof (existential quantifiers)Seeming contradiction of the tertium non datur principle through a logic problem
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classical logic - rules for quantifiers
The Next CEO of Stack OverflowCan second order logic express each (computable) infinitary logic sentence?Logic coursework assistanceSubstitution not clear in logicStuck on a quantifier logic problemAre proofs for many-sorted first order logic shorter than single sorted first order logic?Prove each equivalence by using the rules for semantic equivalenceClassical logic without negation and falsehoodCurious tableaux rules for a curious logicpredicate logic proof (existential quantifiers)Seeming contradiction of the tertium non datur principle through a logic problem
$begingroup$
I have these formulas of CL:
(a) ∀xP(x,x)
(b) ∀x∀y∀z(P(x,y)∧P(y,z) → P(x,z))
(c) ∀x∀y(P(x,y) → ¬P(y,x)
and I have been trying to prove weather (a),(b) ⊨ (c). First I would use ∀l and then my question is, can I apply this rule for every single ∀x (y or z) and then just have many eigenvariables or is there another way?
If anyone knows of exercises similar to this for practice that would be very helpful
logic proof-theory sequent-calculus
$endgroup$
add a comment |
$begingroup$
I have these formulas of CL:
(a) ∀xP(x,x)
(b) ∀x∀y∀z(P(x,y)∧P(y,z) → P(x,z))
(c) ∀x∀y(P(x,y) → ¬P(y,x)
and I have been trying to prove weather (a),(b) ⊨ (c). First I would use ∀l and then my question is, can I apply this rule for every single ∀x (y or z) and then just have many eigenvariables or is there another way?
If anyone knows of exercises similar to this for practice that would be very helpful
logic proof-theory sequent-calculus
$endgroup$
7
$begingroup$
Are you sure that you have to prove that $forall x P(x,x), forall x forall y forall z (P(x,y) land P(y,z) to P(x,z)) vdash forall x forall y (P(x,y) to lnot P(y,x)$ ? If you interpret $P$ as the equality, you can see that such a sequent is not provable.
$endgroup$
– Taroccoesbrocco
Mar 20 at 11:29
$begingroup$
In general, do you already have the completeness theorem for first-order logic at hand? Then it may be useful for any kind of derivability question to consider the corresponding version using the semantic entailment $models$.
$endgroup$
– blub
Mar 20 at 11:40
3
$begingroup$
(a), (b) $vDash$ (c) means that there is no interpretation that satisfies (a) and (b) but falsifies (c). You can check with a simple domain $D = a,b $ such that $P(a,a), P(a,b), P(b,a)$ and $P(b,b)$ all hold. In this case, both premises are TRUE and the conclusion is FALSE.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 12:58
add a comment |
$begingroup$
I have these formulas of CL:
(a) ∀xP(x,x)
(b) ∀x∀y∀z(P(x,y)∧P(y,z) → P(x,z))
(c) ∀x∀y(P(x,y) → ¬P(y,x)
and I have been trying to prove weather (a),(b) ⊨ (c). First I would use ∀l and then my question is, can I apply this rule for every single ∀x (y or z) and then just have many eigenvariables or is there another way?
If anyone knows of exercises similar to this for practice that would be very helpful
logic proof-theory sequent-calculus
$endgroup$
I have these formulas of CL:
(a) ∀xP(x,x)
(b) ∀x∀y∀z(P(x,y)∧P(y,z) → P(x,z))
(c) ∀x∀y(P(x,y) → ¬P(y,x)
and I have been trying to prove weather (a),(b) ⊨ (c). First I would use ∀l and then my question is, can I apply this rule for every single ∀x (y or z) and then just have many eigenvariables or is there another way?
If anyone knows of exercises similar to this for practice that would be very helpful
logic proof-theory sequent-calculus
logic proof-theory sequent-calculus
edited Mar 20 at 14:10
Graham Kemp
87.6k43578
87.6k43578
asked Mar 20 at 11:25
rumetalmIrumetalmI
263
263
7
$begingroup$
Are you sure that you have to prove that $forall x P(x,x), forall x forall y forall z (P(x,y) land P(y,z) to P(x,z)) vdash forall x forall y (P(x,y) to lnot P(y,x)$ ? If you interpret $P$ as the equality, you can see that such a sequent is not provable.
$endgroup$
– Taroccoesbrocco
Mar 20 at 11:29
$begingroup$
In general, do you already have the completeness theorem for first-order logic at hand? Then it may be useful for any kind of derivability question to consider the corresponding version using the semantic entailment $models$.
$endgroup$
– blub
Mar 20 at 11:40
3
$begingroup$
(a), (b) $vDash$ (c) means that there is no interpretation that satisfies (a) and (b) but falsifies (c). You can check with a simple domain $D = a,b $ such that $P(a,a), P(a,b), P(b,a)$ and $P(b,b)$ all hold. In this case, both premises are TRUE and the conclusion is FALSE.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 12:58
add a comment |
7
$begingroup$
Are you sure that you have to prove that $forall x P(x,x), forall x forall y forall z (P(x,y) land P(y,z) to P(x,z)) vdash forall x forall y (P(x,y) to lnot P(y,x)$ ? If you interpret $P$ as the equality, you can see that such a sequent is not provable.
$endgroup$
– Taroccoesbrocco
Mar 20 at 11:29
$begingroup$
In general, do you already have the completeness theorem for first-order logic at hand? Then it may be useful for any kind of derivability question to consider the corresponding version using the semantic entailment $models$.
$endgroup$
– blub
Mar 20 at 11:40
3
$begingroup$
(a), (b) $vDash$ (c) means that there is no interpretation that satisfies (a) and (b) but falsifies (c). You can check with a simple domain $D = a,b $ such that $P(a,a), P(a,b), P(b,a)$ and $P(b,b)$ all hold. In this case, both premises are TRUE and the conclusion is FALSE.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 12:58
7
7
$begingroup$
Are you sure that you have to prove that $forall x P(x,x), forall x forall y forall z (P(x,y) land P(y,z) to P(x,z)) vdash forall x forall y (P(x,y) to lnot P(y,x)$ ? If you interpret $P$ as the equality, you can see that such a sequent is not provable.
$endgroup$
– Taroccoesbrocco
Mar 20 at 11:29
$begingroup$
Are you sure that you have to prove that $forall x P(x,x), forall x forall y forall z (P(x,y) land P(y,z) to P(x,z)) vdash forall x forall y (P(x,y) to lnot P(y,x)$ ? If you interpret $P$ as the equality, you can see that such a sequent is not provable.
$endgroup$
– Taroccoesbrocco
Mar 20 at 11:29
$begingroup$
In general, do you already have the completeness theorem for first-order logic at hand? Then it may be useful for any kind of derivability question to consider the corresponding version using the semantic entailment $models$.
$endgroup$
– blub
Mar 20 at 11:40
$begingroup$
In general, do you already have the completeness theorem for first-order logic at hand? Then it may be useful for any kind of derivability question to consider the corresponding version using the semantic entailment $models$.
$endgroup$
– blub
Mar 20 at 11:40
3
3
$begingroup$
(a), (b) $vDash$ (c) means that there is no interpretation that satisfies (a) and (b) but falsifies (c). You can check with a simple domain $D = a,b $ such that $P(a,a), P(a,b), P(b,a)$ and $P(b,b)$ all hold. In this case, both premises are TRUE and the conclusion is FALSE.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 12:58
$begingroup$
(a), (b) $vDash$ (c) means that there is no interpretation that satisfies (a) and (b) but falsifies (c). You can check with a simple domain $D = a,b $ such that $P(a,a), P(a,b), P(b,a)$ and $P(b,b)$ all hold. In this case, both premises are TRUE and the conclusion is FALSE.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 12:58
add a comment |
0
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oldest
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7
$begingroup$
Are you sure that you have to prove that $forall x P(x,x), forall x forall y forall z (P(x,y) land P(y,z) to P(x,z)) vdash forall x forall y (P(x,y) to lnot P(y,x)$ ? If you interpret $P$ as the equality, you can see that such a sequent is not provable.
$endgroup$
– Taroccoesbrocco
Mar 20 at 11:29
$begingroup$
In general, do you already have the completeness theorem for first-order logic at hand? Then it may be useful for any kind of derivability question to consider the corresponding version using the semantic entailment $models$.
$endgroup$
– blub
Mar 20 at 11:40
3
$begingroup$
(a), (b) $vDash$ (c) means that there is no interpretation that satisfies (a) and (b) but falsifies (c). You can check with a simple domain $D = a,b $ such that $P(a,a), P(a,b), P(b,a)$ and $P(b,b)$ all hold. In this case, both premises are TRUE and the conclusion is FALSE.
$endgroup$
– Mauro ALLEGRANZA
Mar 20 at 12:58