Completeness of the Gentzen Sequent Calculus












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Let $A_1, dots, A_n ,B_1, dots, B_m in mathcal{F}_{¬,lor}$ with $models lnot A_1 lor ldots lor lnot A_n lor B_1 lor ldots lor B_m$ ($mathcal{F}_{¬,lor}$ is the set of formulas built up from propositional variables and the connectives $lnot$ and $lor$). Show that this implies that the sequent $A_1, ldots, A_n vdash_G B_1, ldots, B_m$ is derivable in the Gentzen sequent calculus.
Use induction over the overall number of operators ($lor$ and $lnot$) in $A_1, dots, A_n, B_1, dots, B_m$.



Note: As a special case of the statement above we get: if $models B_1$ then $vdash_G B_1$ is derivable (for $B_1 in mathcal{F}_{lnot,lor}$).



I understand Gentzen sequence calculus. But I am unable to prove this using induction.






logic propositional-calculus proof-theory sequent-calculus







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    $begingroup$


    Let $A_1, dots, A_n ,B_1, dots, B_m in mathcal{F}_{¬,lor}$ with $models lnot A_1 lor ldots lor lnot A_n lor B_1 lor ldots lor B_m$ ($mathcal{F}_{¬,lor}$ is the set of formulas built up from propositional variables and the connectives $lnot$ and $lor$). Show that this implies that the sequent $A_1, ldots, A_n vdash_G B_1, ldots, B_m$ is derivable in the Gentzen sequent calculus.
    Use induction over the overall number of operators ($lor$ and $lnot$) in $A_1, dots, A_n, B_1, dots, B_m$.



    Note: As a special case of the statement above we get: if $models B_1$ then $vdash_G B_1$ is derivable (for $B_1 in mathcal{F}_{lnot,lor}$).



    I understand Gentzen sequence calculus. But I am unable to prove this using induction.






    logic propositional-calculus proof-theory sequent-calculus







    share|cite|improve this question











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      $begingroup$


      Let $A_1, dots, A_n ,B_1, dots, B_m in mathcal{F}_{¬,lor}$ with $models lnot A_1 lor ldots lor lnot A_n lor B_1 lor ldots lor B_m$ ($mathcal{F}_{¬,lor}$ is the set of formulas built up from propositional variables and the connectives $lnot$ and $lor$). Show that this implies that the sequent $A_1, ldots, A_n vdash_G B_1, ldots, B_m$ is derivable in the Gentzen sequent calculus.
      Use induction over the overall number of operators ($lor$ and $lnot$) in $A_1, dots, A_n, B_1, dots, B_m$.



      Note: As a special case of the statement above we get: if $models B_1$ then $vdash_G B_1$ is derivable (for $B_1 in mathcal{F}_{lnot,lor}$).



      I understand Gentzen sequence calculus. But I am unable to prove this using induction.






      logic propositional-calculus proof-theory sequent-calculus







      share|cite|improve this question











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      Let $A_1, dots, A_n ,B_1, dots, B_m in mathcal{F}_{¬,lor}$ with $models lnot A_1 lor ldots lor lnot A_n lor B_1 lor ldots lor B_m$ ($mathcal{F}_{¬,lor}$ is the set of formulas built up from propositional variables and the connectives $lnot$ and $lor$). Show that this implies that the sequent $A_1, ldots, A_n vdash_G B_1, ldots, B_m$ is derivable in the Gentzen sequent calculus.
      Use induction over the overall number of operators ($lor$ and $lnot$) in $A_1, dots, A_n, B_1, dots, B_m$.



      Note: As a special case of the statement above we get: if $models B_1$ then $vdash_G B_1$ is derivable (for $B_1 in mathcal{F}_{lnot,lor}$).



      I understand Gentzen sequence calculus. But I am unable to prove this using induction.






      logic propositional-calculus proof-theory sequent-calculus




      logic propositional-calculus proof-theory sequent-calculus






      share|cite|improve this question















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      edited Jan 23 at 1:08









      Taroccoesbrocco

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      asked Aug 20 '17 at 18:33









      user2397555user2397555

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          The proof of your statement is by induction on the number $k in mathbb{N}$ of occurrences of $lor$ and $lnot$ in the finite sequence of formulas $A_1, dots, A_n, B_1, dots, B_m$ (I refer to the inference rules for Gentzen sequent calculus listed here).



          If $k = 0$ then all $A_i$'s and $B_j$'s are propositional variables.
          Since $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$, there exist $1 leq i leq n$ and $1 leq j leq m$ such that $A_i = B_j$ (otherwise $ lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$ is falsified by the valuation assigning true to all $A_i$'s and false to all $B_j$'s). Therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
          begin{equation}
          dfrac{dfrac{dfrac{}{A_i vdash B_j}text{ax}}{A_i vdash B_1, dots, B_m}text{w}_R text{ ($m-1$ times)}}{A_1, dots, A_n vdash B_1, dots, B_m}text{w}_L text{ ($n-1$ times)}.
          end{equation}



          If $k > 0$ then there are four cases:




          1. There exists $1 leq i leq n$ such that $A_i = lnot A$.
            As $models lnot A_1 lor dots lor lnot lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_{i-1} lor lnot A_{i+1} lor dots lor lnot A_n lor A lor B_1 lor dots lor B_m$.
            By induction hypothesis, $A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
            begin{equation}
            dfrac{overset{vdots}{A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m}}{A_1, dots, lnot A, dots A_n vdash B_1, dots, B_m}lnot_L.
            end{equation}

          2. There exists $1 leq i leq n$ such that $A_i = A lor A'$.
            As $models lnot A_1 lor dots lor lnot (A lor A') lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ and $models lnot A_1 lor dots lor lnot A' lor dots lor lnot A_n lor B_1 lor dots lor B_m$.
            By induction hypothesis, $A_1, dots, A, dots, A_n vdash B_1, dots, B_m$ and $A_1, dots, A', dots, A_n vdash B_1, dots, B_m$ are derivable;
            therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
            begin{equation}
            dfrac{overset{vdots}{A_1, dots, A, dots A_n vdash B_1, dots, B_m} qquad overset{vdots}{A_1, dots, A', dots A_n vdash B_1, dots, B_m}}{A_1, dots, A lor A', dots A_n vdash B_1, dots, B_m}lor_L.
            end{equation}

          3. There exists $1 leq j leq m$ such that $B_j = lnot B$.
            As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor lnot B lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor lnot B lor B_1 lor dots lor B_m$.
            By induction hypothesis, $A_1, dots, A_n, B vdash B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
            begin{equation}
            dfrac{overset{vdots}{A_1, dots, A_n, B vdash B_1, dots, B_m}}{A_1, dots, A_n vdash B_1, dots, lnot B, dots, B_m}lnot_R.
            end{equation}

          4. There exists $1 leq j leq m$ such that $B_j = B lor B'$.
            As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor (B lor B') lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor B lor B' lor B_1 lor dots lor B_m$.
            By induction hypothesis, $A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
            begin{equation}
            dfrac{overset{vdots}{A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m}}{A_1, dots, A_n vdash B_1, dots, B lor B', dots, B_m}lor_R.
            end{equation}






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            $begingroup$

            The proof of your statement is by induction on the number $k in mathbb{N}$ of occurrences of $lor$ and $lnot$ in the finite sequence of formulas $A_1, dots, A_n, B_1, dots, B_m$ (I refer to the inference rules for Gentzen sequent calculus listed here).



            If $k = 0$ then all $A_i$'s and $B_j$'s are propositional variables.
            Since $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$, there exist $1 leq i leq n$ and $1 leq j leq m$ such that $A_i = B_j$ (otherwise $ lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$ is falsified by the valuation assigning true to all $A_i$'s and false to all $B_j$'s). Therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
            begin{equation}
            dfrac{dfrac{dfrac{}{A_i vdash B_j}text{ax}}{A_i vdash B_1, dots, B_m}text{w}_R text{ ($m-1$ times)}}{A_1, dots, A_n vdash B_1, dots, B_m}text{w}_L text{ ($n-1$ times)}.
            end{equation}



            If $k > 0$ then there are four cases:




            1. There exists $1 leq i leq n$ such that $A_i = lnot A$.
              As $models lnot A_1 lor dots lor lnot lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_{i-1} lor lnot A_{i+1} lor dots lor lnot A_n lor A lor B_1 lor dots lor B_m$.
              By induction hypothesis, $A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
              begin{equation}
              dfrac{overset{vdots}{A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m}}{A_1, dots, lnot A, dots A_n vdash B_1, dots, B_m}lnot_L.
              end{equation}

            2. There exists $1 leq i leq n$ such that $A_i = A lor A'$.
              As $models lnot A_1 lor dots lor lnot (A lor A') lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ and $models lnot A_1 lor dots lor lnot A' lor dots lor lnot A_n lor B_1 lor dots lor B_m$.
              By induction hypothesis, $A_1, dots, A, dots, A_n vdash B_1, dots, B_m$ and $A_1, dots, A', dots, A_n vdash B_1, dots, B_m$ are derivable;
              therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
              begin{equation}
              dfrac{overset{vdots}{A_1, dots, A, dots A_n vdash B_1, dots, B_m} qquad overset{vdots}{A_1, dots, A', dots A_n vdash B_1, dots, B_m}}{A_1, dots, A lor A', dots A_n vdash B_1, dots, B_m}lor_L.
              end{equation}

            3. There exists $1 leq j leq m$ such that $B_j = lnot B$.
              As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor lnot B lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor lnot B lor B_1 lor dots lor B_m$.
              By induction hypothesis, $A_1, dots, A_n, B vdash B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
              begin{equation}
              dfrac{overset{vdots}{A_1, dots, A_n, B vdash B_1, dots, B_m}}{A_1, dots, A_n vdash B_1, dots, lnot B, dots, B_m}lnot_R.
              end{equation}

            4. There exists $1 leq j leq m$ such that $B_j = B lor B'$.
              As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor (B lor B') lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor B lor B' lor B_1 lor dots lor B_m$.
              By induction hypothesis, $A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
              begin{equation}
              dfrac{overset{vdots}{A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m}}{A_1, dots, A_n vdash B_1, dots, B lor B', dots, B_m}lor_R.
              end{equation}






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            $endgroup$


















              1












              $begingroup$

              The proof of your statement is by induction on the number $k in mathbb{N}$ of occurrences of $lor$ and $lnot$ in the finite sequence of formulas $A_1, dots, A_n, B_1, dots, B_m$ (I refer to the inference rules for Gentzen sequent calculus listed here).



              If $k = 0$ then all $A_i$'s and $B_j$'s are propositional variables.
              Since $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$, there exist $1 leq i leq n$ and $1 leq j leq m$ such that $A_i = B_j$ (otherwise $ lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$ is falsified by the valuation assigning true to all $A_i$'s and false to all $B_j$'s). Therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
              begin{equation}
              dfrac{dfrac{dfrac{}{A_i vdash B_j}text{ax}}{A_i vdash B_1, dots, B_m}text{w}_R text{ ($m-1$ times)}}{A_1, dots, A_n vdash B_1, dots, B_m}text{w}_L text{ ($n-1$ times)}.
              end{equation}



              If $k > 0$ then there are four cases:




              1. There exists $1 leq i leq n$ such that $A_i = lnot A$.
                As $models lnot A_1 lor dots lor lnot lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_{i-1} lor lnot A_{i+1} lor dots lor lnot A_n lor A lor B_1 lor dots lor B_m$.
                By induction hypothesis, $A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                begin{equation}
                dfrac{overset{vdots}{A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m}}{A_1, dots, lnot A, dots A_n vdash B_1, dots, B_m}lnot_L.
                end{equation}

              2. There exists $1 leq i leq n$ such that $A_i = A lor A'$.
                As $models lnot A_1 lor dots lor lnot (A lor A') lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ and $models lnot A_1 lor dots lor lnot A' lor dots lor lnot A_n lor B_1 lor dots lor B_m$.
                By induction hypothesis, $A_1, dots, A, dots, A_n vdash B_1, dots, B_m$ and $A_1, dots, A', dots, A_n vdash B_1, dots, B_m$ are derivable;
                therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                begin{equation}
                dfrac{overset{vdots}{A_1, dots, A, dots A_n vdash B_1, dots, B_m} qquad overset{vdots}{A_1, dots, A', dots A_n vdash B_1, dots, B_m}}{A_1, dots, A lor A', dots A_n vdash B_1, dots, B_m}lor_L.
                end{equation}

              3. There exists $1 leq j leq m$ such that $B_j = lnot B$.
                As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor lnot B lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor lnot B lor B_1 lor dots lor B_m$.
                By induction hypothesis, $A_1, dots, A_n, B vdash B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                begin{equation}
                dfrac{overset{vdots}{A_1, dots, A_n, B vdash B_1, dots, B_m}}{A_1, dots, A_n vdash B_1, dots, lnot B, dots, B_m}lnot_R.
                end{equation}

              4. There exists $1 leq j leq m$ such that $B_j = B lor B'$.
                As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor (B lor B') lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor B lor B' lor B_1 lor dots lor B_m$.
                By induction hypothesis, $A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                begin{equation}
                dfrac{overset{vdots}{A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m}}{A_1, dots, A_n vdash B_1, dots, B lor B', dots, B_m}lor_R.
                end{equation}






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                $begingroup$

                The proof of your statement is by induction on the number $k in mathbb{N}$ of occurrences of $lor$ and $lnot$ in the finite sequence of formulas $A_1, dots, A_n, B_1, dots, B_m$ (I refer to the inference rules for Gentzen sequent calculus listed here).



                If $k = 0$ then all $A_i$'s and $B_j$'s are propositional variables.
                Since $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$, there exist $1 leq i leq n$ and $1 leq j leq m$ such that $A_i = B_j$ (otherwise $ lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$ is falsified by the valuation assigning true to all $A_i$'s and false to all $B_j$'s). Therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                begin{equation}
                dfrac{dfrac{dfrac{}{A_i vdash B_j}text{ax}}{A_i vdash B_1, dots, B_m}text{w}_R text{ ($m-1$ times)}}{A_1, dots, A_n vdash B_1, dots, B_m}text{w}_L text{ ($n-1$ times)}.
                end{equation}



                If $k > 0$ then there are four cases:




                1. There exists $1 leq i leq n$ such that $A_i = lnot A$.
                  As $models lnot A_1 lor dots lor lnot lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_{i-1} lor lnot A_{i+1} lor dots lor lnot A_n lor A lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m}}{A_1, dots, lnot A, dots A_n vdash B_1, dots, B_m}lnot_L.
                  end{equation}

                2. There exists $1 leq i leq n$ such that $A_i = A lor A'$.
                  As $models lnot A_1 lor dots lor lnot (A lor A') lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ and $models lnot A_1 lor dots lor lnot A' lor dots lor lnot A_n lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A, dots, A_n vdash B_1, dots, B_m$ and $A_1, dots, A', dots, A_n vdash B_1, dots, B_m$ are derivable;
                  therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A, dots A_n vdash B_1, dots, B_m} qquad overset{vdots}{A_1, dots, A', dots A_n vdash B_1, dots, B_m}}{A_1, dots, A lor A', dots A_n vdash B_1, dots, B_m}lor_L.
                  end{equation}

                3. There exists $1 leq j leq m$ such that $B_j = lnot B$.
                  As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor lnot B lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor lnot B lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A_n, B vdash B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A_n, B vdash B_1, dots, B_m}}{A_1, dots, A_n vdash B_1, dots, lnot B, dots, B_m}lnot_R.
                  end{equation}

                4. There exists $1 leq j leq m$ such that $B_j = B lor B'$.
                  As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor (B lor B') lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor B lor B' lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m}}{A_1, dots, A_n vdash B_1, dots, B lor B', dots, B_m}lor_R.
                  end{equation}






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                $endgroup$



                The proof of your statement is by induction on the number $k in mathbb{N}$ of occurrences of $lor$ and $lnot$ in the finite sequence of formulas $A_1, dots, A_n, B_1, dots, B_m$ (I refer to the inference rules for Gentzen sequent calculus listed here).



                If $k = 0$ then all $A_i$'s and $B_j$'s are propositional variables.
                Since $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$, there exist $1 leq i leq n$ and $1 leq j leq m$ such that $A_i = B_j$ (otherwise $ lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor B_m$ is falsified by the valuation assigning true to all $A_i$'s and false to all $B_j$'s). Therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                begin{equation}
                dfrac{dfrac{dfrac{}{A_i vdash B_j}text{ax}}{A_i vdash B_1, dots, B_m}text{w}_R text{ ($m-1$ times)}}{A_1, dots, A_n vdash B_1, dots, B_m}text{w}_L text{ ($n-1$ times)}.
                end{equation}



                If $k > 0$ then there are four cases:




                1. There exists $1 leq i leq n$ such that $A_i = lnot A$.
                  As $models lnot A_1 lor dots lor lnot lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_{i-1} lor lnot A_{i+1} lor dots lor lnot A_n lor A lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A_{i-1}, A_{i+1}, dots A_n vdash A, B_1, dots, B_m}}{A_1, dots, lnot A, dots A_n vdash B_1, dots, B_m}lnot_L.
                  end{equation}

                2. There exists $1 leq i leq n$ such that $A_i = A lor A'$.
                  As $models lnot A_1 lor dots lor lnot (A lor A') lor dots lor lnot A_n lor B_1 lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A lor dots lor lnot A_n lor B_1 lor dots lor B_m$ and $models lnot A_1 lor dots lor lnot A' lor dots lor lnot A_n lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A, dots, A_n vdash B_1, dots, B_m$ and $A_1, dots, A', dots, A_n vdash B_1, dots, B_m$ are derivable;
                  therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A, dots A_n vdash B_1, dots, B_m} qquad overset{vdots}{A_1, dots, A', dots A_n vdash B_1, dots, B_m}}{A_1, dots, A lor A', dots A_n vdash B_1, dots, B_m}lor_L.
                  end{equation}

                3. There exists $1 leq j leq m$ such that $B_j = lnot B$.
                  As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor lnot B lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor lnot B lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A_n, B vdash B_1, dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A_n, B vdash B_1, dots, B_m}}{A_1, dots, A_n vdash B_1, dots, lnot B, dots, B_m}lnot_R.
                  end{equation}

                4. There exists $1 leq j leq m$ such that $B_j = B lor B'$.
                  As $models lnot A_1 lor dots lor lnot A_n lor B_1 lor dots lor (B lor B') lor dots lor B_m$ by hypothesis, then $models lnot A_1 lor dots lor lnot A_n lor B lor B' lor B_1 lor dots lor B_m$.
                  By induction hypothesis, $A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m$ is derivable; therefore, $A_1, dots, A_n vdash B_1, dots, B_m$ is derivable, for instance by the following derivation
                  begin{equation}
                  dfrac{overset{vdots}{A_1, dots, A_n vdash B_1, dots, B, B', dots, B_m}}{A_1, dots, A_n vdash B_1, dots, B lor B', dots, B_m}lor_R.
                  end{equation}







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                answered Jan 23 at 1:06









                TaroccoesbroccoTaroccoesbrocco

                5,64271840




                5,64271840






























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