Assuming $Sigma vdash eta$, what is the deduction for $Sigma vdash negeta rightarrow eta$?
$begingroup$
Assuming $Sigma vdash eta$, what is the deduction for $Sigma
vdash negeta rightarrow eta$?
I understand that $Sigma vdash eta rightarrow Sigma cupnegeta vdash eta$, but I'm trying to specifically find the derivation for $Sigma
vdash negeta rightarrow eta$.
I can't figure out how to do this. I know a deduction is a sequence of logical axioms or non-logical axioms from $Sigma$ or by a rule of inference but any sequence I try doesn't seem to work.
Anyone have any ideas?
The rules of inference I am using are:
Type PC: If $Gamma$ is a finite set of $L$ formulas and $phi$ is an $L$ formula and $phi$ is a propositional consequence of $Gamma$, then $(Gamma, phi)$ is a rule of inference of type PC.
Type QR: Suppose $x$ is a variable that is not free in $psi$. Then $({psi rightarrow phi}, psi rightarrow (forall xphi)), ({phi rightarrow psi}, (exists xphi) rightarrow psi)$
The logical axioms I am using are:
E1: $x=x$ for each variable $x$
E2: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow f(x_1, x_2,, dots, x_n) = f(y_1, y_2, dots, y_n)$
E3: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow (R(x_1, x_2,, dots, x_n) rightarrow R(y_1, y_2, dots, y_n))$
Q1: If $t$ is substitutable for $x$ in $phi$, then $(forall x phi) rightarrow phi_t^x$
Q2: If $t$ is substitutable for $x$ in $phi$, then $phi_t^x rightarrow (exists x phi)$
logic
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
$endgroup$
add a comment |
$begingroup$
Assuming $Sigma vdash eta$, what is the deduction for $Sigma
vdash negeta rightarrow eta$?
I understand that $Sigma vdash eta rightarrow Sigma cupnegeta vdash eta$, but I'm trying to specifically find the derivation for $Sigma
vdash negeta rightarrow eta$.
I can't figure out how to do this. I know a deduction is a sequence of logical axioms or non-logical axioms from $Sigma$ or by a rule of inference but any sequence I try doesn't seem to work.
Anyone have any ideas?
The rules of inference I am using are:
Type PC: If $Gamma$ is a finite set of $L$ formulas and $phi$ is an $L$ formula and $phi$ is a propositional consequence of $Gamma$, then $(Gamma, phi)$ is a rule of inference of type PC.
Type QR: Suppose $x$ is a variable that is not free in $psi$. Then $({psi rightarrow phi}, psi rightarrow (forall xphi)), ({phi rightarrow psi}, (exists xphi) rightarrow psi)$
The logical axioms I am using are:
E1: $x=x$ for each variable $x$
E2: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow f(x_1, x_2,, dots, x_n) = f(y_1, y_2, dots, y_n)$
E3: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow (R(x_1, x_2,, dots, x_n) rightarrow R(y_1, y_2, dots, y_n))$
Q1: If $t$ is substitutable for $x$ in $phi$, then $(forall x phi) rightarrow phi_t^x$
Q2: If $t$ is substitutable for $x$ in $phi$, then $phi_t^x rightarrow (exists x phi)$
logic
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
$endgroup$
$begingroup$
What are your logical axioms and rules of inference? Without knowing this, it is impossible to give you a meaningful answer.
$endgroup$
– Henning Makholm
Dec 30 '18 at 14:35
$begingroup$
@HenningMakholm I have edited the question to include the logical axioms and rules of inference.
$endgroup$
– Oliver G
Dec 30 '18 at 15:03
add a comment |
$begingroup$
Assuming $Sigma vdash eta$, what is the deduction for $Sigma
vdash negeta rightarrow eta$?
I understand that $Sigma vdash eta rightarrow Sigma cupnegeta vdash eta$, but I'm trying to specifically find the derivation for $Sigma
vdash negeta rightarrow eta$.
I can't figure out how to do this. I know a deduction is a sequence of logical axioms or non-logical axioms from $Sigma$ or by a rule of inference but any sequence I try doesn't seem to work.
Anyone have any ideas?
The rules of inference I am using are:
Type PC: If $Gamma$ is a finite set of $L$ formulas and $phi$ is an $L$ formula and $phi$ is a propositional consequence of $Gamma$, then $(Gamma, phi)$ is a rule of inference of type PC.
Type QR: Suppose $x$ is a variable that is not free in $psi$. Then $({psi rightarrow phi}, psi rightarrow (forall xphi)), ({phi rightarrow psi}, (exists xphi) rightarrow psi)$
The logical axioms I am using are:
E1: $x=x$ for each variable $x$
E2: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow f(x_1, x_2,, dots, x_n) = f(y_1, y_2, dots, y_n)$
E3: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow (R(x_1, x_2,, dots, x_n) rightarrow R(y_1, y_2, dots, y_n))$
Q1: If $t$ is substitutable for $x$ in $phi$, then $(forall x phi) rightarrow phi_t^x$
Q2: If $t$ is substitutable for $x$ in $phi$, then $phi_t^x rightarrow (exists x phi)$
logic
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
$endgroup$
Assuming $Sigma vdash eta$, what is the deduction for $Sigma
vdash negeta rightarrow eta$?
I understand that $Sigma vdash eta rightarrow Sigma cupnegeta vdash eta$, but I'm trying to specifically find the derivation for $Sigma
vdash negeta rightarrow eta$.
I can't figure out how to do this. I know a deduction is a sequence of logical axioms or non-logical axioms from $Sigma$ or by a rule of inference but any sequence I try doesn't seem to work.
Anyone have any ideas?
The rules of inference I am using are:
Type PC: If $Gamma$ is a finite set of $L$ formulas and $phi$ is an $L$ formula and $phi$ is a propositional consequence of $Gamma$, then $(Gamma, phi)$ is a rule of inference of type PC.
Type QR: Suppose $x$ is a variable that is not free in $psi$. Then $({psi rightarrow phi}, psi rightarrow (forall xphi)), ({phi rightarrow psi}, (exists xphi) rightarrow psi)$
The logical axioms I am using are:
E1: $x=x$ for each variable $x$
E2: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow f(x_1, x_2,, dots, x_n) = f(y_1, y_2, dots, y_n)$
E3: $[(x_1=y_1) land (x_2=y_2) land dots (x_n = y_n)] rightarrow (R(x_1, x_2,, dots, x_n) rightarrow R(y_1, y_2, dots, y_n))$
Q1: If $t$ is substitutable for $x$ in $phi$, then $(forall x phi) rightarrow phi_t^x$
Q2: If $t$ is substitutable for $x$ in $phi$, then $phi_t^x rightarrow (exists x phi)$
logic
logic
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
edited Dec 30 '18 at 15:03
Oliver G
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
asked Dec 30 '18 at 14:28
Oliver GOliver G
1,4551632
1,4551632
$begingroup$
What are your logical axioms and rules of inference? Without knowing this, it is impossible to give you a meaningful answer.
$endgroup$
– Henning Makholm
Dec 30 '18 at 14:35
$begingroup$
@HenningMakholm I have edited the question to include the logical axioms and rules of inference.
$endgroup$
– Oliver G
Dec 30 '18 at 15:03
add a comment |
$begingroup$
What are your logical axioms and rules of inference? Without knowing this, it is impossible to give you a meaningful answer.
$endgroup$
– Henning Makholm
Dec 30 '18 at 14:35
$begingroup$
@HenningMakholm I have edited the question to include the logical axioms and rules of inference.
$endgroup$
– Oliver G
Dec 30 '18 at 15:03
$begingroup$
What are your logical axioms and rules of inference? Without knowing this, it is impossible to give you a meaningful answer.
$endgroup$
– Henning Makholm
Dec 30 '18 at 14:35
$begingroup$
What are your logical axioms and rules of inference? Without knowing this, it is impossible to give you a meaningful answer.
$endgroup$
– Henning Makholm
Dec 30 '18 at 14:35
$begingroup$
@HenningMakholm I have edited the question to include the logical axioms and rules of inference.
$endgroup$
– Oliver G
Dec 30 '18 at 15:03
$begingroup$
@HenningMakholm I have edited the question to include the logical axioms and rules of inference.
$endgroup$
– Oliver G
Dec 30 '18 at 15:03
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$negetatoeta$ is a propositional consequence of $eta$.
So if you have a derivation of $Sigmavdash eta$, appending a single PC step to it will give you a derivation of $Sigmavdashnegetatoeta$.
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
$endgroup$
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
oldest
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$begingroup$
$negetatoeta$ is a propositional consequence of $eta$.
So if you have a derivation of $Sigmavdash eta$, appending a single PC step to it will give you a derivation of $Sigmavdashnegetatoeta$.
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
$endgroup$
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
add a comment |
$begingroup$
$negetatoeta$ is a propositional consequence of $eta$.
So if you have a derivation of $Sigmavdash eta$, appending a single PC step to it will give you a derivation of $Sigmavdashnegetatoeta$.
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
$endgroup$
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
add a comment |
$begingroup$
$negetatoeta$ is a propositional consequence of $eta$.
So if you have a derivation of $Sigmavdash eta$, appending a single PC step to it will give you a derivation of $Sigmavdashnegetatoeta$.
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
$endgroup$
$negetatoeta$ is a propositional consequence of $eta$.
So if you have a derivation of $Sigmavdash eta$, appending a single PC step to it will give you a derivation of $Sigmavdashnegetatoeta$.
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
answered Dec 30 '18 at 15:11
Henning MakholmHenning Makholm
242k17308551
242k17308551
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
add a comment |
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
I see now. $eta rightarrow (neg eta rightarrow eta)$ is the appropriate step since $eta$ is the only thing being assumed we can then claim $neg eta rightarrow eta$ is in the deduction.
$endgroup$
– Oliver G
Dec 30 '18 at 15:14
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
$begingroup$
Otherwise known as weakening
$endgroup$
– Agnishom Chattopadhyay
Dec 30 '18 at 15:17
add a comment |
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$begingroup$
What are your logical axioms and rules of inference? Without knowing this, it is impossible to give you a meaningful answer.
$endgroup$
– Henning Makholm
Dec 30 '18 at 14:35
$begingroup$
@HenningMakholm I have edited the question to include the logical axioms and rules of inference.
$endgroup$
– Oliver G
Dec 30 '18 at 15:03