show that for every consistent theory there is a complete consistent theory












0














Let $mathcal{L}$ be any language of predicate logic, $Sigma_0$ a consistent theory in $mathcal{L}$. Let P be the set of all consistent theories $Sigma supseteq Sigma_0$ in $mathcal{L}$.
With the relation $subseteq$ P becomes partial ordered.



I'm trying to show that

1) every chain in P (i.e., every through $subseteq$ total ordered subset K$subseteq$ P) is limited. (I.e. for every chain $K subseteq P$ there is $Sigma^{ast} in P$ such that for all $Sigma in K$ $Sigma subseteq Sigma^{ast}$ holds.
The suggestion ist to check, if the constructed set $Sigma^{ast}$ really is in P and to not forget that not every chain is orderisomorphic (I hope that's the correct English word for it) to $(mathbb{N}, <)$;



2) if $Sigma in P_{mathcal{L}}$ is maximal (i.e. there's no $Sigma' supseteq Sigma$ in P), then $Sigma$ is complete;



3) conclude with the Lemma of Zorn that for every consistent theory there is a complete consistent theory, which is superset of it.



My problem lies within point 1). Neither do I see intuitively how this holds nor do I have an idea of how to proof it.
3) Will be easy once 1) and 2) are shown, I think.



So I'm looking forward to your ideas for 1) and 2).






first-order-logic predicate-logic proof-theory formal-proofs







share|cite|improve this question





























    0














    Let $mathcal{L}$ be any language of predicate logic, $Sigma_0$ a consistent theory in $mathcal{L}$. Let P be the set of all consistent theories $Sigma supseteq Sigma_0$ in $mathcal{L}$.
    With the relation $subseteq$ P becomes partial ordered.



    I'm trying to show that

    1) every chain in P (i.e., every through $subseteq$ total ordered subset K$subseteq$ P) is limited. (I.e. for every chain $K subseteq P$ there is $Sigma^{ast} in P$ such that for all $Sigma in K$ $Sigma subseteq Sigma^{ast}$ holds.
    The suggestion ist to check, if the constructed set $Sigma^{ast}$ really is in P and to not forget that not every chain is orderisomorphic (I hope that's the correct English word for it) to $(mathbb{N}, <)$;



    2) if $Sigma in P_{mathcal{L}}$ is maximal (i.e. there's no $Sigma' supseteq Sigma$ in P), then $Sigma$ is complete;



    3) conclude with the Lemma of Zorn that for every consistent theory there is a complete consistent theory, which is superset of it.



    My problem lies within point 1). Neither do I see intuitively how this holds nor do I have an idea of how to proof it.
    3) Will be easy once 1) and 2) are shown, I think.



    So I'm looking forward to your ideas for 1) and 2).






    first-order-logic predicate-logic proof-theory formal-proofs







    share|cite|improve this question



























      0












      0








      0







      Let $mathcal{L}$ be any language of predicate logic, $Sigma_0$ a consistent theory in $mathcal{L}$. Let P be the set of all consistent theories $Sigma supseteq Sigma_0$ in $mathcal{L}$.
      With the relation $subseteq$ P becomes partial ordered.



      I'm trying to show that

      1) every chain in P (i.e., every through $subseteq$ total ordered subset K$subseteq$ P) is limited. (I.e. for every chain $K subseteq P$ there is $Sigma^{ast} in P$ such that for all $Sigma in K$ $Sigma subseteq Sigma^{ast}$ holds.
      The suggestion ist to check, if the constructed set $Sigma^{ast}$ really is in P and to not forget that not every chain is orderisomorphic (I hope that's the correct English word for it) to $(mathbb{N}, <)$;



      2) if $Sigma in P_{mathcal{L}}$ is maximal (i.e. there's no $Sigma' supseteq Sigma$ in P), then $Sigma$ is complete;



      3) conclude with the Lemma of Zorn that for every consistent theory there is a complete consistent theory, which is superset of it.



      My problem lies within point 1). Neither do I see intuitively how this holds nor do I have an idea of how to proof it.
      3) Will be easy once 1) and 2) are shown, I think.



      So I'm looking forward to your ideas for 1) and 2).






      first-order-logic predicate-logic proof-theory formal-proofs







      share|cite|improve this question















      Let $mathcal{L}$ be any language of predicate logic, $Sigma_0$ a consistent theory in $mathcal{L}$. Let P be the set of all consistent theories $Sigma supseteq Sigma_0$ in $mathcal{L}$.
      With the relation $subseteq$ P becomes partial ordered.



      I'm trying to show that

      1) every chain in P (i.e., every through $subseteq$ total ordered subset K$subseteq$ P) is limited. (I.e. for every chain $K subseteq P$ there is $Sigma^{ast} in P$ such that for all $Sigma in K$ $Sigma subseteq Sigma^{ast}$ holds.
      The suggestion ist to check, if the constructed set $Sigma^{ast}$ really is in P and to not forget that not every chain is orderisomorphic (I hope that's the correct English word for it) to $(mathbb{N}, <)$;



      2) if $Sigma in P_{mathcal{L}}$ is maximal (i.e. there's no $Sigma' supseteq Sigma$ in P), then $Sigma$ is complete;



      3) conclude with the Lemma of Zorn that for every consistent theory there is a complete consistent theory, which is superset of it.



      My problem lies within point 1). Neither do I see intuitively how this holds nor do I have an idea of how to proof it.
      3) Will be easy once 1) and 2) are shown, I think.



      So I'm looking forward to your ideas for 1) and 2).






      first-order-logic predicate-logic proof-theory formal-proofs




      first-order-logic predicate-logic proof-theory formal-proofs






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 at 22:19

























      asked Nov 24 at 17:47









      Studentu

      988




      988






















          1 Answer
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          For part one, you want to take $Sigma^*$ to be the union of the chain $K$. You need to show it is a consistent theory extending $Sigma_0.$ That it extends $Sigma_0$ is obvious. If it were inconsistent, the inconsistency would come from some finite number of sentences $varphi_1,ldots varphi_nin Sigma^*$. For each $varphi_i,$ choose a $Sigma_iin K $ such that $varphi_iin Sigma_i.$ Then take $Sigma$ to be the maximum of $Sigma_1,ldots, Sigma_n.$ Then $Sigmain K,$ and is inconsistent, contradicting the fact that $K$ only contains consistent sets of statements.



          I’m not sure what to make of the suggestion that $K$ is not necessarily order isomorphic to $mathbb N.$ I guess maybe people in the past have written the above proof making a tacit assumption that it is and they want to warn you against that.



          For 2, show that if $Sigma$ is not complete, then there is a sentence $varphinotin Sigma$ such that $Sigmacup{varphi}$ is consistent.






          share|cite|improve this answer























          • Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
            – Studentu
            Nov 25 at 4:22












          • 3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
            – Studentu
            Nov 25 at 4:23








          • 1




            For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
            – spaceisdarkgreen
            Nov 25 at 6:11






          • 1




            The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
            – spaceisdarkgreen
            Nov 25 at 19:34






          • 1




            @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
            – spaceisdarkgreen
            Nov 25 at 21:16













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          For part one, you want to take $Sigma^*$ to be the union of the chain $K$. You need to show it is a consistent theory extending $Sigma_0.$ That it extends $Sigma_0$ is obvious. If it were inconsistent, the inconsistency would come from some finite number of sentences $varphi_1,ldots varphi_nin Sigma^*$. For each $varphi_i,$ choose a $Sigma_iin K $ such that $varphi_iin Sigma_i.$ Then take $Sigma$ to be the maximum of $Sigma_1,ldots, Sigma_n.$ Then $Sigmain K,$ and is inconsistent, contradicting the fact that $K$ only contains consistent sets of statements.



          I’m not sure what to make of the suggestion that $K$ is not necessarily order isomorphic to $mathbb N.$ I guess maybe people in the past have written the above proof making a tacit assumption that it is and they want to warn you against that.



          For 2, show that if $Sigma$ is not complete, then there is a sentence $varphinotin Sigma$ such that $Sigmacup{varphi}$ is consistent.






          share|cite|improve this answer























          • Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
            – Studentu
            Nov 25 at 4:22












          • 3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
            – Studentu
            Nov 25 at 4:23








          • 1




            For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
            – spaceisdarkgreen
            Nov 25 at 6:11






          • 1




            The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
            – spaceisdarkgreen
            Nov 25 at 19:34






          • 1




            @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
            – spaceisdarkgreen
            Nov 25 at 21:16


















          1














          For part one, you want to take $Sigma^*$ to be the union of the chain $K$. You need to show it is a consistent theory extending $Sigma_0.$ That it extends $Sigma_0$ is obvious. If it were inconsistent, the inconsistency would come from some finite number of sentences $varphi_1,ldots varphi_nin Sigma^*$. For each $varphi_i,$ choose a $Sigma_iin K $ such that $varphi_iin Sigma_i.$ Then take $Sigma$ to be the maximum of $Sigma_1,ldots, Sigma_n.$ Then $Sigmain K,$ and is inconsistent, contradicting the fact that $K$ only contains consistent sets of statements.



          I’m not sure what to make of the suggestion that $K$ is not necessarily order isomorphic to $mathbb N.$ I guess maybe people in the past have written the above proof making a tacit assumption that it is and they want to warn you against that.



          For 2, show that if $Sigma$ is not complete, then there is a sentence $varphinotin Sigma$ such that $Sigmacup{varphi}$ is consistent.






          share|cite|improve this answer























          • Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
            – Studentu
            Nov 25 at 4:22












          • 3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
            – Studentu
            Nov 25 at 4:23








          • 1




            For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
            – spaceisdarkgreen
            Nov 25 at 6:11






          • 1




            The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
            – spaceisdarkgreen
            Nov 25 at 19:34






          • 1




            @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
            – spaceisdarkgreen
            Nov 25 at 21:16
















          1












          1








          1






          For part one, you want to take $Sigma^*$ to be the union of the chain $K$. You need to show it is a consistent theory extending $Sigma_0.$ That it extends $Sigma_0$ is obvious. If it were inconsistent, the inconsistency would come from some finite number of sentences $varphi_1,ldots varphi_nin Sigma^*$. For each $varphi_i,$ choose a $Sigma_iin K $ such that $varphi_iin Sigma_i.$ Then take $Sigma$ to be the maximum of $Sigma_1,ldots, Sigma_n.$ Then $Sigmain K,$ and is inconsistent, contradicting the fact that $K$ only contains consistent sets of statements.



          I’m not sure what to make of the suggestion that $K$ is not necessarily order isomorphic to $mathbb N.$ I guess maybe people in the past have written the above proof making a tacit assumption that it is and they want to warn you against that.



          For 2, show that if $Sigma$ is not complete, then there is a sentence $varphinotin Sigma$ such that $Sigmacup{varphi}$ is consistent.






          share|cite|improve this answer














          For part one, you want to take $Sigma^*$ to be the union of the chain $K$. You need to show it is a consistent theory extending $Sigma_0.$ That it extends $Sigma_0$ is obvious. If it were inconsistent, the inconsistency would come from some finite number of sentences $varphi_1,ldots varphi_nin Sigma^*$. For each $varphi_i,$ choose a $Sigma_iin K $ such that $varphi_iin Sigma_i.$ Then take $Sigma$ to be the maximum of $Sigma_1,ldots, Sigma_n.$ Then $Sigmain K,$ and is inconsistent, contradicting the fact that $K$ only contains consistent sets of statements.



          I’m not sure what to make of the suggestion that $K$ is not necessarily order isomorphic to $mathbb N.$ I guess maybe people in the past have written the above proof making a tacit assumption that it is and they want to warn you against that.



          For 2, show that if $Sigma$ is not complete, then there is a sentence $varphinotin Sigma$ such that $Sigmacup{varphi}$ is consistent.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 6:12

























          answered Nov 24 at 18:27









          spaceisdarkgreen

          32.3k21753




          32.3k21753












          • Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
            – Studentu
            Nov 25 at 4:22












          • 3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
            – Studentu
            Nov 25 at 4:23








          • 1




            For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
            – spaceisdarkgreen
            Nov 25 at 6:11






          • 1




            The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
            – spaceisdarkgreen
            Nov 25 at 19:34






          • 1




            @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
            – spaceisdarkgreen
            Nov 25 at 21:16




















          • Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
            – Studentu
            Nov 25 at 4:22












          • 3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
            – Studentu
            Nov 25 at 4:23








          • 1




            For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
            – spaceisdarkgreen
            Nov 25 at 6:11






          • 1




            The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
            – spaceisdarkgreen
            Nov 25 at 19:34






          • 1




            @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
            – spaceisdarkgreen
            Nov 25 at 21:16


















          Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
          – Studentu
          Nov 25 at 4:22






          Thank you! I'd completely forgotten about the trick to take the union! For 2), we assume that $Sigma$ is not complete. So there is $phi$ in $mathcal{L}$ such that it doesn't hold that either $Sigma vdash Phi$ or $Sigma vdash neg phi$. We consider $Sigma':=Sigma cup phi$. We do this with all $psi$ which cause $Sigma$ to not be complete until we have the theory $tilde{Sigma}$. Then $tilde{Sigma}supset Sigma$ and $tilde{Sigma} in P$ in contradiction to the maximality of $Sigma$. But I don't see how to explain the $tilde{Sigma}$ is consistent?
          – Studentu
          Nov 25 at 4:22














          3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
          – Studentu
          Nov 25 at 4:23






          3) With the Lemma of Zorn we get that there is at least one maximal element in P. Due to 2), this element is also complete. But I don't get how to explain that 3) follows from this. Intuitively, it is clear to me, but if we take the maximal element $Sigma^{ast}$ for example, it doesn't have to hold that $Sigma_0 in Sigma^{ast}$, because it could be another maximal element which $Sigma_0$ is a subset of.
          – Studentu
          Nov 25 at 4:23






          1




          1




          For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
          – spaceisdarkgreen
          Nov 25 at 6:11




          For 2 you only need to show that if $Sigma nvdash phi$ and $Sigma nvdash lnotphi,$ then either $Sigmacup{phi}$ or $Sigmacup{lnotphi}$ are consistent (actually, both are). This shows that if $Sigmain P$ is incomplete, then it is not maximal in $P,$ which is what you need. For 3, yes it does have to hold that $Sigma_0subseteq Sigma^*$ since all $Sigma in P$ have $Sigma_0subseteqSigma.$
          – spaceisdarkgreen
          Nov 25 at 6:11




          1




          1




          The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
          – spaceisdarkgreen
          Nov 25 at 19:34




          The general theorem you can show it that $Sigmacup {phi}$ is consistent if and only if $Sigma nvdash lnotphi .$ For the direction you need, assume $Sigmacup {phi}$ is inconsistent then show $Sigma vdash lnot phi.$ You only need to use very general facts about the system. For instance by the deduction theorem, if $Sigmacup {phi}vdash psi,$ and $Sigmacup {phi}vdash lnotpsi,$ then $ Sigma vdash phito psi$ and $Sigma vdash phito lnot psi.$ Can you show from there that $Sigma vdash lnotphi$?
          – spaceisdarkgreen
          Nov 25 at 19:34




          1




          1




          @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
          – spaceisdarkgreen
          Nov 25 at 21:16






          @Studentu Some of that is on the right track, but you need $Sigma vdash lnotphi,$ not just $Sigmacup{phi}vdash lnot phi.$ If you have the $bot,$ symbol, you might just do $ Sigmacup{phi}vdashbot,$ and then use the deduction theorem to get $Sigma vdash phito bot.$ How close this is to $Sigmavdash lnot phi$ depends on the system you're using (in some systems it's the same thing by definition).
          – spaceisdarkgreen
          Nov 25 at 21:16




















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