Help with Hilbert Calculus The 2019 Stack Overflow Developer Survey Results Are InWhy is this true? $(exists x)(P(x) Rightarrow (forall y) P(y))$Why not ban nested quantifiers over the same variable?Proof of Drinker paradoxHow to show that $vdash (forall x beta to alpha) leftrightarrow exists x (beta to alpha)$?Formal “Hilbert-Style” Proof of a relatively simple statementHelp me prove this principle with other Hilbert system principlesHelp needed with axiomatic deductionsPredicate calculus (formal deduction vs resolution)Hilbert-calculus, formal proofHilbert Calculus, Formal Proof ConverseWhich theories are consistent?Associativity of 'or' and 'and' in Hilbert Ackerman SystemRelationship between sequent calculus and Hilbert systems, natural deduction, etcHilbert Calculus Derivation
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Help with Hilbert Calculus
The 2019 Stack Overflow Developer Survey Results Are InWhy is this true? $(exists x)(P(x) Rightarrow (forall y) P(y))$Why not ban nested quantifiers over the same variable?Proof of Drinker paradoxHow to show that $vdash (forall x beta to alpha) leftrightarrow exists x (beta to alpha)$?Formal “Hilbert-Style” Proof of a relatively simple statementHelp me prove this principle with other Hilbert system principlesHelp needed with axiomatic deductionsPredicate calculus (formal deduction vs resolution)Hilbert-calculus, formal proofHilbert Calculus, Formal Proof ConverseWhich theories are consistent?Associativity of 'or' and 'and' in Hilbert Ackerman SystemRelationship between sequent calculus and Hilbert systems, natural deduction, etcHilbert Calculus Derivation
$begingroup$
Can u help show that this is a theorem? $ (∀x_1 (∃x_2 (p(x_1, x_2) ⇒ (∀x_2 p(x_1, x_2)))));$
I was trying to use the deduction theorem but i hit a wall. Can u help me out using derivatives and Hilbert Calculus?
logic first-order-logic predicate-logic hilbert-calculus
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
$endgroup$
add a comment |
$begingroup$
Can u help show that this is a theorem? $ (∀x_1 (∃x_2 (p(x_1, x_2) ⇒ (∀x_2 p(x_1, x_2)))));$
I was trying to use the deduction theorem but i hit a wall. Can u help me out using derivatives and Hilbert Calculus?
logic first-order-logic predicate-logic hilbert-calculus
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
$endgroup$
$begingroup$
It is good that you used parenthesis to disambiguate the expression, but they way you placed them, your second $x_2$ is over writing the first $x_2$, I think that is might be a typographical error.
$endgroup$
– DanielV
Mar 23 at 18:12
$begingroup$
@DanielV: No, that's how it's supposed to go. The body of the $forall x_1$ is an instance of the Drinker paradox.
$endgroup$
– Henning Makholm
Mar 23 at 18:14
1
$begingroup$
@HenningMakholm It still should be $forall x_3(p(x_1,,x_3))$ (say), though.
$endgroup$
– J.G.
Mar 23 at 18:18
1
$begingroup$
See the post Proof of Drinker paradox as well as the post Why is this true? (∃x)(P(x)⇒(∀y)P(y))
$endgroup$
– Mauro ALLEGRANZA
Mar 23 at 18:29
1
$begingroup$
@J.G.: You could also write that. But most formalizations of first-order logic do allow a quantifier to re-bind a variable that is already bound by an enclosing quantifier -- for reasons discussed in this question.
$endgroup$
– Henning Makholm
Mar 23 at 18:30
add a comment |
$begingroup$
Can u help show that this is a theorem? $ (∀x_1 (∃x_2 (p(x_1, x_2) ⇒ (∀x_2 p(x_1, x_2)))));$
I was trying to use the deduction theorem but i hit a wall. Can u help me out using derivatives and Hilbert Calculus?
logic first-order-logic predicate-logic hilbert-calculus
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
$endgroup$
Can u help show that this is a theorem? $ (∀x_1 (∃x_2 (p(x_1, x_2) ⇒ (∀x_2 p(x_1, x_2)))));$
I was trying to use the deduction theorem but i hit a wall. Can u help me out using derivatives and Hilbert Calculus?
logic first-order-logic predicate-logic hilbert-calculus
logic first-order-logic predicate-logic hilbert-calculus
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
edited Mar 24 at 14:26
Mauro ALLEGRANZA
67.8k449117
67.8k449117
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
asked Mar 23 at 17:38
Pedro SantosPedro Santos
17010
17010
$begingroup$
It is good that you used parenthesis to disambiguate the expression, but they way you placed them, your second $x_2$ is over writing the first $x_2$, I think that is might be a typographical error.
$endgroup$
– DanielV
Mar 23 at 18:12
$begingroup$
@DanielV: No, that's how it's supposed to go. The body of the $forall x_1$ is an instance of the Drinker paradox.
$endgroup$
– Henning Makholm
Mar 23 at 18:14
1
$begingroup$
@HenningMakholm It still should be $forall x_3(p(x_1,,x_3))$ (say), though.
$endgroup$
– J.G.
Mar 23 at 18:18
1
$begingroup$
See the post Proof of Drinker paradox as well as the post Why is this true? (∃x)(P(x)⇒(∀y)P(y))
$endgroup$
– Mauro ALLEGRANZA
Mar 23 at 18:29
1
$begingroup$
@J.G.: You could also write that. But most formalizations of first-order logic do allow a quantifier to re-bind a variable that is already bound by an enclosing quantifier -- for reasons discussed in this question.
$endgroup$
– Henning Makholm
Mar 23 at 18:30
add a comment |
$begingroup$
It is good that you used parenthesis to disambiguate the expression, but they way you placed them, your second $x_2$ is over writing the first $x_2$, I think that is might be a typographical error.
$endgroup$
– DanielV
Mar 23 at 18:12
$begingroup$
@DanielV: No, that's how it's supposed to go. The body of the $forall x_1$ is an instance of the Drinker paradox.
$endgroup$
– Henning Makholm
Mar 23 at 18:14
1
$begingroup$
@HenningMakholm It still should be $forall x_3(p(x_1,,x_3))$ (say), though.
$endgroup$
– J.G.
Mar 23 at 18:18
1
$begingroup$
See the post Proof of Drinker paradox as well as the post Why is this true? (∃x)(P(x)⇒(∀y)P(y))
$endgroup$
– Mauro ALLEGRANZA
Mar 23 at 18:29
1
$begingroup$
@J.G.: You could also write that. But most formalizations of first-order logic do allow a quantifier to re-bind a variable that is already bound by an enclosing quantifier -- for reasons discussed in this question.
$endgroup$
– Henning Makholm
Mar 23 at 18:30
$begingroup$
It is good that you used parenthesis to disambiguate the expression, but they way you placed them, your second $x_2$ is over writing the first $x_2$, I think that is might be a typographical error.
$endgroup$
– DanielV
Mar 23 at 18:12
$begingroup$
It is good that you used parenthesis to disambiguate the expression, but they way you placed them, your second $x_2$ is over writing the first $x_2$, I think that is might be a typographical error.
$endgroup$
– DanielV
Mar 23 at 18:12
$begingroup$
@DanielV: No, that's how it's supposed to go. The body of the $forall x_1$ is an instance of the Drinker paradox.
$endgroup$
– Henning Makholm
Mar 23 at 18:14
$begingroup$
@DanielV: No, that's how it's supposed to go. The body of the $forall x_1$ is an instance of the Drinker paradox.
$endgroup$
– Henning Makholm
Mar 23 at 18:14
1
1
$begingroup$
@HenningMakholm It still should be $forall x_3(p(x_1,,x_3))$ (say), though.
$endgroup$
– J.G.
Mar 23 at 18:18
$begingroup$
@HenningMakholm It still should be $forall x_3(p(x_1,,x_3))$ (say), though.
$endgroup$
– J.G.
Mar 23 at 18:18
1
1
$begingroup$
See the post Proof of Drinker paradox as well as the post Why is this true? (∃x)(P(x)⇒(∀y)P(y))
$endgroup$
– Mauro ALLEGRANZA
Mar 23 at 18:29
$begingroup$
See the post Proof of Drinker paradox as well as the post Why is this true? (∃x)(P(x)⇒(∀y)P(y))
$endgroup$
– Mauro ALLEGRANZA
Mar 23 at 18:29
1
1
$begingroup$
@J.G.: You could also write that. But most formalizations of first-order logic do allow a quantifier to re-bind a variable that is already bound by an enclosing quantifier -- for reasons discussed in this question.
$endgroup$
– Henning Makholm
Mar 23 at 18:30
$begingroup$
@J.G.: You could also write that. But most formalizations of first-order logic do allow a quantifier to re-bind a variable that is already bound by an enclosing quantifier -- for reasons discussed in this question.
$endgroup$
– Henning Makholm
Mar 23 at 18:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
"English" answer:
Fix any $x_1,,x_2$. If $p(x_1,,x_2)$ is false, the required implication is vacuously true. If $x_2$ cannot be chosen so that $p(x_1,,x_2)$ is false, our choice of $x_1$ has obtained $forall x_2 (p(x_1,,x_2))$, so again the implication succeeds.
HC answer:
$$exists x_2(neg p(x_1,,x_2))implies(p(x_1,,x_2)implies forall x_3 (p(x_1,,x_3)))$$
$$notexists x_2(neg p(x_1,,x_2))impliesforall x_3(p(x_1,,x_3)),,implies(p(x_1,,x_2)impliesforall x_3(p(x_1,,x_3)))$$
Then use $(qimplies r)land (neg qimplies r)implies r$.
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
$endgroup$
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
add a comment |
$begingroup$
Hint
In this post you can find an Hilbert-style proof of :
$⊢(∀xβ → α) ↔ ∃x(β → α)$, provided that $x$ is not free in $alpha$.
We have to consider :
$∀x_2p(x_1,x_2) to ∀x_2p(x_1,x_2)$;
it is an instance of the propositional tautology : $vdash A to A$, and thus is a theorem.
Now we apply the equivalence above to it, due to the fact that $x_2$ is not free in $∀x_2p(x_1,x_2)$ to get :
$vdash ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
The last step is obtained with Generalization:
$vdash ∀x_1 ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
$endgroup$
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
$begingroup$
"English" answer:
Fix any $x_1,,x_2$. If $p(x_1,,x_2)$ is false, the required implication is vacuously true. If $x_2$ cannot be chosen so that $p(x_1,,x_2)$ is false, our choice of $x_1$ has obtained $forall x_2 (p(x_1,,x_2))$, so again the implication succeeds.
HC answer:
$$exists x_2(neg p(x_1,,x_2))implies(p(x_1,,x_2)implies forall x_3 (p(x_1,,x_3)))$$
$$notexists x_2(neg p(x_1,,x_2))impliesforall x_3(p(x_1,,x_3)),,implies(p(x_1,,x_2)impliesforall x_3(p(x_1,,x_3)))$$
Then use $(qimplies r)land (neg qimplies r)implies r$.
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
$endgroup$
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
add a comment |
$begingroup$
"English" answer:
Fix any $x_1,,x_2$. If $p(x_1,,x_2)$ is false, the required implication is vacuously true. If $x_2$ cannot be chosen so that $p(x_1,,x_2)$ is false, our choice of $x_1$ has obtained $forall x_2 (p(x_1,,x_2))$, so again the implication succeeds.
HC answer:
$$exists x_2(neg p(x_1,,x_2))implies(p(x_1,,x_2)implies forall x_3 (p(x_1,,x_3)))$$
$$notexists x_2(neg p(x_1,,x_2))impliesforall x_3(p(x_1,,x_3)),,implies(p(x_1,,x_2)impliesforall x_3(p(x_1,,x_3)))$$
Then use $(qimplies r)land (neg qimplies r)implies r$.
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
$endgroup$
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
add a comment |
$begingroup$
"English" answer:
Fix any $x_1,,x_2$. If $p(x_1,,x_2)$ is false, the required implication is vacuously true. If $x_2$ cannot be chosen so that $p(x_1,,x_2)$ is false, our choice of $x_1$ has obtained $forall x_2 (p(x_1,,x_2))$, so again the implication succeeds.
HC answer:
$$exists x_2(neg p(x_1,,x_2))implies(p(x_1,,x_2)implies forall x_3 (p(x_1,,x_3)))$$
$$notexists x_2(neg p(x_1,,x_2))impliesforall x_3(p(x_1,,x_3)),,implies(p(x_1,,x_2)impliesforall x_3(p(x_1,,x_3)))$$
Then use $(qimplies r)land (neg qimplies r)implies r$.
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
$endgroup$
"English" answer:
Fix any $x_1,,x_2$. If $p(x_1,,x_2)$ is false, the required implication is vacuously true. If $x_2$ cannot be chosen so that $p(x_1,,x_2)$ is false, our choice of $x_1$ has obtained $forall x_2 (p(x_1,,x_2))$, so again the implication succeeds.
HC answer:
$$exists x_2(neg p(x_1,,x_2))implies(p(x_1,,x_2)implies forall x_3 (p(x_1,,x_3)))$$
$$notexists x_2(neg p(x_1,,x_2))impliesforall x_3(p(x_1,,x_3)),,implies(p(x_1,,x_2)impliesforall x_3(p(x_1,,x_3)))$$
Then use $(qimplies r)land (neg qimplies r)implies r$.
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
edited Mar 23 at 18:42
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
answered Mar 23 at 17:46
J.G.J.G.
33.2k23252
33.2k23252
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
add a comment |
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
Yeah Thanks , but i was trying to do it using a sequece of derivatives with hilbert calculus.
$endgroup$
– Pedro Santos
Mar 23 at 18:18
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
$begingroup$
@PedroSantos I hope my edit is a little bit more useful.
$endgroup$
– J.G.
Mar 23 at 18:42
add a comment |
$begingroup$
Hint
In this post you can find an Hilbert-style proof of :
$⊢(∀xβ → α) ↔ ∃x(β → α)$, provided that $x$ is not free in $alpha$.
We have to consider :
$∀x_2p(x_1,x_2) to ∀x_2p(x_1,x_2)$;
it is an instance of the propositional tautology : $vdash A to A$, and thus is a theorem.
Now we apply the equivalence above to it, due to the fact that $x_2$ is not free in $∀x_2p(x_1,x_2)$ to get :
$vdash ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
The last step is obtained with Generalization:
$vdash ∀x_1 ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
$endgroup$
add a comment |
$begingroup$
Hint
In this post you can find an Hilbert-style proof of :
$⊢(∀xβ → α) ↔ ∃x(β → α)$, provided that $x$ is not free in $alpha$.
We have to consider :
$∀x_2p(x_1,x_2) to ∀x_2p(x_1,x_2)$;
it is an instance of the propositional tautology : $vdash A to A$, and thus is a theorem.
Now we apply the equivalence above to it, due to the fact that $x_2$ is not free in $∀x_2p(x_1,x_2)$ to get :
$vdash ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
The last step is obtained with Generalization:
$vdash ∀x_1 ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
$endgroup$
add a comment |
$begingroup$
Hint
In this post you can find an Hilbert-style proof of :
$⊢(∀xβ → α) ↔ ∃x(β → α)$, provided that $x$ is not free in $alpha$.
We have to consider :
$∀x_2p(x_1,x_2) to ∀x_2p(x_1,x_2)$;
it is an instance of the propositional tautology : $vdash A to A$, and thus is a theorem.
Now we apply the equivalence above to it, due to the fact that $x_2$ is not free in $∀x_2p(x_1,x_2)$ to get :
$vdash ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
The last step is obtained with Generalization:
$vdash ∀x_1 ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
$endgroup$
Hint
In this post you can find an Hilbert-style proof of :
$⊢(∀xβ → α) ↔ ∃x(β → α)$, provided that $x$ is not free in $alpha$.
We have to consider :
$∀x_2p(x_1,x_2) to ∀x_2p(x_1,x_2)$;
it is an instance of the propositional tautology : $vdash A to A$, and thus is a theorem.
Now we apply the equivalence above to it, due to the fact that $x_2$ is not free in $∀x_2p(x_1,x_2)$ to get :
$vdash ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
The last step is obtained with Generalization:
$vdash ∀x_1 ∃x_2 (p(x_1,x_2) to ∀x_2p(x_1,x_2))$.
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
answered Mar 24 at 14:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.8k449117
67.8k449117
add a comment |
add a comment |
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$begingroup$
It is good that you used parenthesis to disambiguate the expression, but they way you placed them, your second $x_2$ is over writing the first $x_2$, I think that is might be a typographical error.
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– DanielV
Mar 23 at 18:12
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@DanielV: No, that's how it's supposed to go. The body of the $forall x_1$ is an instance of the Drinker paradox.
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– Henning Makholm
Mar 23 at 18:14
1
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@HenningMakholm It still should be $forall x_3(p(x_1,,x_3))$ (say), though.
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– J.G.
Mar 23 at 18:18
1
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See the post Proof of Drinker paradox as well as the post Why is this true? (∃x)(P(x)⇒(∀y)P(y))
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– Mauro ALLEGRANZA
Mar 23 at 18:29
1
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@J.G.: You could also write that. But most formalizations of first-order logic do allow a quantifier to re-bind a variable that is already bound by an enclosing quantifier -- for reasons discussed in this question.
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– Henning Makholm
Mar 23 at 18:30