Mixing
sequent calculus for first order logic
Clash Royale CLAN TAG#URR8PPP
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I've just started learning sequent calculus. Now I'm trying to prove the formula below:
$$ exists x (P → Q) ⊨ P → forall x Q $$
My approach to the problem:
$$ underline_⊢exists x (P → Q) , _⊣ P → forall x Q $$
$$ underline_⊢exists x (P → Q) , _⊢P, _⊣forall x Q $$
$$ underline_⊢P(y) → Q(y), _⊢P, _⊣forall x Q $$
$$ underline_⊣P(y), _⊢P, _⊣forall x Q (1) _⊢Q(y), _⊢P, _⊣forall x Q (2) $$
In the second case we have $ _⊣forall x Q $, thus $ _⊣ Q(y) $ - contradiction (because of $_⊢Q(y)$) .
In the first one we only have $ _⊣P(y) $ and $ _⊢P $, so there is a counterexample for the formula:
$$ P(x)=T, textif x = y and F text otherwise $$
$$ Q(x)=F forall x $$
Am I right? (An unquantified $P$ is a bit embarrasing for me)
predicate-logic proof-theory sequent-calculus
asked Jul 18 at 13:35
mikha koltjia
183
add a comment |Â
up vote
2
down vote
favorite
I've just started learning sequent calculus. Now I'm trying to prove the formula below:
$$ exists x (P → Q) ⊨ P → forall x Q $$
My approach to the problem:
$$ underline_⊢exists x (P → Q) , _⊣ P → forall x Q $$
$$ underline_⊢exists x (P → Q) , _⊢P, _⊣forall x Q $$
$$ underline_⊢P(y) → Q(y), _⊢P, _⊣forall x Q $$
$$ underline_⊣P(y), _⊢P, _⊣forall x Q (1) _⊢Q(y), _⊢P, _⊣forall x Q (2) $$
In the second case we have $ _⊣forall x Q $, thus $ _⊣ Q(y) $ - contradiction (because of $_⊢Q(y)$) .
In the first one we only have $ _⊣P(y) $ and $ _⊢P $, so there is a counterexample for the formula:
$$ P(x)=T, textif x = y and F text otherwise $$
$$ Q(x)=F forall x $$
Am I right? (An unquantified $P$ is a bit embarrasing for me)
predicate-logic proof-theory sequent-calculus
asked Jul 18 at 13:35
mikha koltjia
183
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've just started learning sequent calculus. Now I'm trying to prove the formula below:
$$ exists x (P → Q) ⊨ P → forall x Q $$
My approach to the problem:
$$ underline_⊢exists x (P → Q) , _⊣ P → forall x Q $$
$$ underline_⊢exists x (P → Q) , _⊢P, _⊣forall x Q $$
$$ underline_⊢P(y) → Q(y), _⊢P, _⊣forall x Q $$
$$ underline_⊣P(y), _⊢P, _⊣forall x Q (1) _⊢Q(y), _⊢P, _⊣forall x Q (2) $$
In the second case we have $ _⊣forall x Q $, thus $ _⊣ Q(y) $ - contradiction (because of $_⊢Q(y)$) .
In the first one we only have $ _⊣P(y) $ and $ _⊢P $, so there is a counterexample for the formula:
$$ P(x)=T, textif x = y and F text otherwise $$
$$ Q(x)=F forall x $$
Am I right? (An unquantified $P$ is a bit embarrasing for me)
predicate-logic proof-theory sequent-calculus
asked Jul 18 at 13:35
mikha koltjia
183
I've just started learning sequent calculus. Now I'm trying to prove the formula below:
$$ exists x (P → Q) ⊨ P → forall x Q $$
My approach to the problem:
$$ underline_⊢exists x (P → Q) , _⊣ P → forall x Q $$
$$ underline_⊢exists x (P → Q) , _⊢P, _⊣forall x Q $$
$$ underline_⊢P(y) → Q(y), _⊢P, _⊣forall x Q $$
$$ underline_⊣P(y), _⊢P, _⊣forall x Q (1) _⊢Q(y), _⊢P, _⊣forall x Q (2) $$
In the second case we have $ _⊣forall x Q $, thus $ _⊣ Q(y) $ - contradiction (because of $_⊢Q(y)$) .
In the first one we only have $ _⊣P(y) $ and $ _⊢P $, so there is a counterexample for the formula:
$$ P(x)=T, textif x = y and F text otherwise $$
$$ Q(x)=F forall x $$
Am I right? (An unquantified $P$ is a bit embarrasing for me)
predicate-logic proof-theory sequent-calculus
asked Jul 18 at 13:35
mikha koltjia
183
asked Jul 18 at 13:35
mikha koltjia
183
asked Jul 18 at 13:35
mikha koltjia
183
asked Jul 18 at 13:35
mikha koltjia
183
183
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Long comment
Quite a strange way of writing sequents...
Having said that, and assuming that $x$ is free in $P$, the answer is :
$exists x (Px to Qx) nvDash Px to forall x Qx$.
Assume a domain with one Black ball and one White ball and interpret $P$ with "... is Black" and $Q$ with "... is White".
We have that $Px$ is FALSE for the White ball, and thus the conditional $Px to Qx$ is TRUE of it.
Thus, in this interpretation, the formula $exists x (Px to Qx)$ is TRUE.
But $P$ is TRUE of the Black ball and $forall x Qx$ is FALSE (not every ball is White).
Thus, in this interpretation, the formula $Px to forall x Qx$ is FALSE.
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Long comment
Quite a strange way of writing sequents...
Having said that, and assuming that $x$ is free in $P$, the answer is :
$exists x (Px to Qx) nvDash Px to forall x Qx$.
Assume a domain with one Black ball and one White ball and interpret $P$ with "... is Black" and $Q$ with "... is White".
We have that $Px$ is FALSE for the White ball, and thus the conditional $Px to Qx$ is TRUE of it.
Thus, in this interpretation, the formula $exists x (Px to Qx)$ is TRUE.
But $P$ is TRUE of the Black ball and $forall x Qx$ is FALSE (not every ball is White).
Thus, in this interpretation, the formula $Px to forall x Qx$ is FALSE.
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
add a comment |Â
up vote
1
down vote
accepted
Long comment
Quite a strange way of writing sequents...
Having said that, and assuming that $x$ is free in $P$, the answer is :
$exists x (Px to Qx) nvDash Px to forall x Qx$.
Assume a domain with one Black ball and one White ball and interpret $P$ with "... is Black" and $Q$ with "... is White".
We have that $Px$ is FALSE for the White ball, and thus the conditional $Px to Qx$ is TRUE of it.
Thus, in this interpretation, the formula $exists x (Px to Qx)$ is TRUE.
But $P$ is TRUE of the Black ball and $forall x Qx$ is FALSE (not every ball is White).
Thus, in this interpretation, the formula $Px to forall x Qx$ is FALSE.
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Long comment
Quite a strange way of writing sequents...
Having said that, and assuming that $x$ is free in $P$, the answer is :
$exists x (Px to Qx) nvDash Px to forall x Qx$.
Assume a domain with one Black ball and one White ball and interpret $P$ with "... is Black" and $Q$ with "... is White".
We have that $Px$ is FALSE for the White ball, and thus the conditional $Px to Qx$ is TRUE of it.
Thus, in this interpretation, the formula $exists x (Px to Qx)$ is TRUE.
But $P$ is TRUE of the Black ball and $forall x Qx$ is FALSE (not every ball is White).
Thus, in this interpretation, the formula $Px to forall x Qx$ is FALSE.
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
Long comment
Quite a strange way of writing sequents...
Having said that, and assuming that $x$ is free in $P$, the answer is :
$exists x (Px to Qx) nvDash Px to forall x Qx$.
Assume a domain with one Black ball and one White ball and interpret $P$ with "... is Black" and $Q$ with "... is White".
We have that $Px$ is FALSE for the White ball, and thus the conditional $Px to Qx$ is TRUE of it.
Thus, in this interpretation, the formula $exists x (Px to Qx)$ is TRUE.
But $P$ is TRUE of the Black ball and $forall x Qx$ is FALSE (not every ball is White).
Thus, in this interpretation, the formula $Px to forall x Qx$ is FALSE.
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
answered Jul 18 at 13:49
Mauro ALLEGRANZA
60.7k346105
60.7k346105
add a comment |Â
add a comment |Â
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