Question about $Gamma_infty^M$ deriving a very specific wff












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In p.69 of Tourlakis's mathematical logic book.



He pulls a very specific theorem seemingly out of thin air.



$Gamma_infty^M vdash exists_x(f bar n ... = x)$



I'm not sure how this is derived as there are no theorems about the restricted theory and function symbols.



The closest thing I found was it derives some general $exists_x(A)$ but can $A = f bar n ... = x$ in every case? Is this the correct train of thought?



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logic first-order-logic model-theory







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    0












    $begingroup$


    In p.69 of Tourlakis's mathematical logic book.



    He pulls a very specific theorem seemingly out of thin air.



    $Gamma_infty^M vdash exists_x(f bar n ... = x)$



    I'm not sure how this is derived as there are no theorems about the restricted theory and function symbols.



    The closest thing I found was it derives some general $exists_x(A)$ but can $A = f bar n ... = x$ in every case? Is this the correct train of thought?



    enter image description here






    logic first-order-logic model-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      In p.69 of Tourlakis's mathematical logic book.



      He pulls a very specific theorem seemingly out of thin air.



      $Gamma_infty^M vdash exists_x(f bar n ... = x)$



      I'm not sure how this is derived as there are no theorems about the restricted theory and function symbols.



      The closest thing I found was it derives some general $exists_x(A)$ but can $A = f bar n ... = x$ in every case? Is this the correct train of thought?



      enter image description here






      logic first-order-logic model-theory







      share|cite|improve this question









      $endgroup$




      In p.69 of Tourlakis's mathematical logic book.



      He pulls a very specific theorem seemingly out of thin air.



      $Gamma_infty^M vdash exists_x(f bar n ... = x)$



      I'm not sure how this is derived as there are no theorems about the restricted theory and function symbols.



      The closest thing I found was it derives some general $exists_x(A)$ but can $A = f bar n ... = x$ in every case? Is this the correct train of thought?



      enter image description here






      logic first-order-logic model-theory




      logic first-order-logic model-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 11 at 11:52









      japseowjapseow

      865




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          2 Answers
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          1












          $begingroup$

          The context is the "Henkin construction" of the model needed for the proof of the Completeness Theorem.



          At page 65 you can see the definition of the set $Gamma_{infty}$ of formulas.



          At page 67 there is the definition of its "restriction" $Gamma_{infty}^M$.



          $M$ is defined as a subset of $mathbb N$ :




          $M = { f (n) : n in mathbb N }$




          where $f(n) = text { the smallest } m text { such that } m ∼ n$ [using Definition I.5.25 of page 66 for the equivalence relation $∼$].



          Consider now page 69 :




          Let next $f_k$ be a function letter of arity $k$,




          this is not the fucntion $f$ above. The new one is a symbol of the language.




          and let $n_1,ldots, n_k$ be an input for $f_k^I$. What is the appropriate output? [I.e. what is $f_k^I (n_1,ldots, n_k)$ ? ]



          Well, first observe that $Gamma_{infty}^M vdash (∃x) f_k(overline {n_1},ldots, overline {n_k}) = x$




          Why? By axioms for equality : $x=x$ [Ax3., page 34], followed by Corollary I.4.12 (Substitution of Terms) : $vdash f_k(overline {n_1},ldots, overline {n_k})=f_k(overline {n_1},ldots, overline {n_k})$.



          Finally apply Ax2. : $mathcal A[x ← t] to (∃x) mathcal A text { for any term } t$, to get :





          $vdash (∃x)f_k(overline {n_1},ldots, overline {n_k})=x$.








          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
            $endgroup$
            – japseow
            Jan 13 at 11:53



















          4












          $begingroup$

          For any language, any theory $Gamma$, and any terms $t_1,...,t_k$ not containing the variable $x$, any function symbol $f$ of arity $k$, you have $Gamma vdash (exists x) ft_1...t_k = x$.



          That's for the simple reason that you have $Gamma vdash ft_1...t_k = t$ with $t=ft_1...t_k$ and so you can use Existential introduction on $t$.



          Note that Existential Introduction is the rule that (under the right conditions) states that if $Gamma vdash P(t)$ for some term $t$, then $Gammavdash (exists x) P(x)$. This rule makes sense : if you know that $P(t)$ is true for a specific $t$, then you know that it's true for some $x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah yes this is exactly what I was missing. Thanks!
            $endgroup$
            – japseow
            Jan 13 at 11:54











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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

          oldest

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          1












          $begingroup$

          The context is the "Henkin construction" of the model needed for the proof of the Completeness Theorem.



          At page 65 you can see the definition of the set $Gamma_{infty}$ of formulas.



          At page 67 there is the definition of its "restriction" $Gamma_{infty}^M$.



          $M$ is defined as a subset of $mathbb N$ :




          $M = { f (n) : n in mathbb N }$




          where $f(n) = text { the smallest } m text { such that } m ∼ n$ [using Definition I.5.25 of page 66 for the equivalence relation $∼$].



          Consider now page 69 :




          Let next $f_k$ be a function letter of arity $k$,




          this is not the fucntion $f$ above. The new one is a symbol of the language.




          and let $n_1,ldots, n_k$ be an input for $f_k^I$. What is the appropriate output? [I.e. what is $f_k^I (n_1,ldots, n_k)$ ? ]



          Well, first observe that $Gamma_{infty}^M vdash (∃x) f_k(overline {n_1},ldots, overline {n_k}) = x$




          Why? By axioms for equality : $x=x$ [Ax3., page 34], followed by Corollary I.4.12 (Substitution of Terms) : $vdash f_k(overline {n_1},ldots, overline {n_k})=f_k(overline {n_1},ldots, overline {n_k})$.



          Finally apply Ax2. : $mathcal A[x ← t] to (∃x) mathcal A text { for any term } t$, to get :





          $vdash (∃x)f_k(overline {n_1},ldots, overline {n_k})=x$.








          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
            $endgroup$
            – japseow
            Jan 13 at 11:53
















          1












          $begingroup$

          The context is the "Henkin construction" of the model needed for the proof of the Completeness Theorem.



          At page 65 you can see the definition of the set $Gamma_{infty}$ of formulas.



          At page 67 there is the definition of its "restriction" $Gamma_{infty}^M$.



          $M$ is defined as a subset of $mathbb N$ :




          $M = { f (n) : n in mathbb N }$




          where $f(n) = text { the smallest } m text { such that } m ∼ n$ [using Definition I.5.25 of page 66 for the equivalence relation $∼$].



          Consider now page 69 :




          Let next $f_k$ be a function letter of arity $k$,




          this is not the fucntion $f$ above. The new one is a symbol of the language.




          and let $n_1,ldots, n_k$ be an input for $f_k^I$. What is the appropriate output? [I.e. what is $f_k^I (n_1,ldots, n_k)$ ? ]



          Well, first observe that $Gamma_{infty}^M vdash (∃x) f_k(overline {n_1},ldots, overline {n_k}) = x$




          Why? By axioms for equality : $x=x$ [Ax3., page 34], followed by Corollary I.4.12 (Substitution of Terms) : $vdash f_k(overline {n_1},ldots, overline {n_k})=f_k(overline {n_1},ldots, overline {n_k})$.



          Finally apply Ax2. : $mathcal A[x ← t] to (∃x) mathcal A text { for any term } t$, to get :





          $vdash (∃x)f_k(overline {n_1},ldots, overline {n_k})=x$.








          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
            $endgroup$
            – japseow
            Jan 13 at 11:53














          1












          1








          1





          $begingroup$

          The context is the "Henkin construction" of the model needed for the proof of the Completeness Theorem.



          At page 65 you can see the definition of the set $Gamma_{infty}$ of formulas.



          At page 67 there is the definition of its "restriction" $Gamma_{infty}^M$.



          $M$ is defined as a subset of $mathbb N$ :




          $M = { f (n) : n in mathbb N }$




          where $f(n) = text { the smallest } m text { such that } m ∼ n$ [using Definition I.5.25 of page 66 for the equivalence relation $∼$].



          Consider now page 69 :




          Let next $f_k$ be a function letter of arity $k$,




          this is not the fucntion $f$ above. The new one is a symbol of the language.




          and let $n_1,ldots, n_k$ be an input for $f_k^I$. What is the appropriate output? [I.e. what is $f_k^I (n_1,ldots, n_k)$ ? ]



          Well, first observe that $Gamma_{infty}^M vdash (∃x) f_k(overline {n_1},ldots, overline {n_k}) = x$




          Why? By axioms for equality : $x=x$ [Ax3., page 34], followed by Corollary I.4.12 (Substitution of Terms) : $vdash f_k(overline {n_1},ldots, overline {n_k})=f_k(overline {n_1},ldots, overline {n_k})$.



          Finally apply Ax2. : $mathcal A[x ← t] to (∃x) mathcal A text { for any term } t$, to get :





          $vdash (∃x)f_k(overline {n_1},ldots, overline {n_k})=x$.








          share|cite|improve this answer











          $endgroup$



          The context is the "Henkin construction" of the model needed for the proof of the Completeness Theorem.



          At page 65 you can see the definition of the set $Gamma_{infty}$ of formulas.



          At page 67 there is the definition of its "restriction" $Gamma_{infty}^M$.



          $M$ is defined as a subset of $mathbb N$ :




          $M = { f (n) : n in mathbb N }$




          where $f(n) = text { the smallest } m text { such that } m ∼ n$ [using Definition I.5.25 of page 66 for the equivalence relation $∼$].



          Consider now page 69 :




          Let next $f_k$ be a function letter of arity $k$,




          this is not the fucntion $f$ above. The new one is a symbol of the language.




          and let $n_1,ldots, n_k$ be an input for $f_k^I$. What is the appropriate output? [I.e. what is $f_k^I (n_1,ldots, n_k)$ ? ]



          Well, first observe that $Gamma_{infty}^M vdash (∃x) f_k(overline {n_1},ldots, overline {n_k}) = x$




          Why? By axioms for equality : $x=x$ [Ax3., page 34], followed by Corollary I.4.12 (Substitution of Terms) : $vdash f_k(overline {n_1},ldots, overline {n_k})=f_k(overline {n_1},ldots, overline {n_k})$.



          Finally apply Ax2. : $mathcal A[x ← t] to (∃x) mathcal A text { for any term } t$, to get :





          $vdash (∃x)f_k(overline {n_1},ldots, overline {n_k})=x$.









          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 15:56

























          answered Jan 11 at 12:17









          Mauro ALLEGRANZAMauro ALLEGRANZA

          66.9k449115




          66.9k449115












          • $begingroup$
            This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
            $endgroup$
            – japseow
            Jan 13 at 11:53


















          • $begingroup$
            This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
            $endgroup$
            – japseow
            Jan 13 at 11:53
















          $begingroup$
          This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
          $endgroup$
          – japseow
          Jan 13 at 11:53




          $begingroup$
          This answer and Max's answer is pretty much the same, but I would pick this as the answer because it is more formally put.
          $endgroup$
          – japseow
          Jan 13 at 11:53











          4












          $begingroup$

          For any language, any theory $Gamma$, and any terms $t_1,...,t_k$ not containing the variable $x$, any function symbol $f$ of arity $k$, you have $Gamma vdash (exists x) ft_1...t_k = x$.



          That's for the simple reason that you have $Gamma vdash ft_1...t_k = t$ with $t=ft_1...t_k$ and so you can use Existential introduction on $t$.



          Note that Existential Introduction is the rule that (under the right conditions) states that if $Gamma vdash P(t)$ for some term $t$, then $Gammavdash (exists x) P(x)$. This rule makes sense : if you know that $P(t)$ is true for a specific $t$, then you know that it's true for some $x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah yes this is exactly what I was missing. Thanks!
            $endgroup$
            – japseow
            Jan 13 at 11:54
















          4












          $begingroup$

          For any language, any theory $Gamma$, and any terms $t_1,...,t_k$ not containing the variable $x$, any function symbol $f$ of arity $k$, you have $Gamma vdash (exists x) ft_1...t_k = x$.



          That's for the simple reason that you have $Gamma vdash ft_1...t_k = t$ with $t=ft_1...t_k$ and so you can use Existential introduction on $t$.



          Note that Existential Introduction is the rule that (under the right conditions) states that if $Gamma vdash P(t)$ for some term $t$, then $Gammavdash (exists x) P(x)$. This rule makes sense : if you know that $P(t)$ is true for a specific $t$, then you know that it's true for some $x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah yes this is exactly what I was missing. Thanks!
            $endgroup$
            – japseow
            Jan 13 at 11:54














          4












          4








          4





          $begingroup$

          For any language, any theory $Gamma$, and any terms $t_1,...,t_k$ not containing the variable $x$, any function symbol $f$ of arity $k$, you have $Gamma vdash (exists x) ft_1...t_k = x$.



          That's for the simple reason that you have $Gamma vdash ft_1...t_k = t$ with $t=ft_1...t_k$ and so you can use Existential introduction on $t$.



          Note that Existential Introduction is the rule that (under the right conditions) states that if $Gamma vdash P(t)$ for some term $t$, then $Gammavdash (exists x) P(x)$. This rule makes sense : if you know that $P(t)$ is true for a specific $t$, then you know that it's true for some $x$.






          share|cite|improve this answer









          $endgroup$



          For any language, any theory $Gamma$, and any terms $t_1,...,t_k$ not containing the variable $x$, any function symbol $f$ of arity $k$, you have $Gamma vdash (exists x) ft_1...t_k = x$.



          That's for the simple reason that you have $Gamma vdash ft_1...t_k = t$ with $t=ft_1...t_k$ and so you can use Existential introduction on $t$.



          Note that Existential Introduction is the rule that (under the right conditions) states that if $Gamma vdash P(t)$ for some term $t$, then $Gammavdash (exists x) P(x)$. This rule makes sense : if you know that $P(t)$ is true for a specific $t$, then you know that it's true for some $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 12:14









          MaxMax

          15.1k11143




          15.1k11143












          • $begingroup$
            Ah yes this is exactly what I was missing. Thanks!
            $endgroup$
            – japseow
            Jan 13 at 11:54


















          • $begingroup$
            Ah yes this is exactly what I was missing. Thanks!
            $endgroup$
            – japseow
            Jan 13 at 11:54
















          $begingroup$
          Ah yes this is exactly what I was missing. Thanks!
          $endgroup$
          – japseow
          Jan 13 at 11:54




          $begingroup$
          Ah yes this is exactly what I was missing. Thanks!
          $endgroup$
          – japseow
          Jan 13 at 11:54


















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