Prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology












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Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.



Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.










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  • Is it propositional logic? How is $models$ defined?
    – Berci
    Dec 1 '18 at 14:28










  • @Berci I added my definition.
    – heaig
    Dec 1 '18 at 14:39






  • 1




    Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
    – Mauro ALLEGRANZA
    Dec 1 '18 at 14:39
















1














Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.



Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.










share|cite|improve this question
























  • Is it propositional logic? How is $models$ defined?
    – Berci
    Dec 1 '18 at 14:28










  • @Berci I added my definition.
    – heaig
    Dec 1 '18 at 14:39






  • 1




    Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
    – Mauro ALLEGRANZA
    Dec 1 '18 at 14:39














1












1








1







Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.



Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.










share|cite|improve this question















Assume that $H,F$ are formulas. How to prove that ${H_1,dots,H_k} models F$ if and only if $(H_1 land dots land H_k) rightarrow F$ is tautology. My intuition is that's right, but I don't know how to prove this in a clean and precise manner.



Definition $models$: $ H models F$ means that every interpretation suitable for both $H$ and $F$, which is a model for $H$ is also a model for $F$.







logic






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edited Dec 1 '18 at 14:39

























asked Dec 1 '18 at 14:23









heaig

62




62












  • Is it propositional logic? How is $models$ defined?
    – Berci
    Dec 1 '18 at 14:28










  • @Berci I added my definition.
    – heaig
    Dec 1 '18 at 14:39






  • 1




    Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
    – Mauro ALLEGRANZA
    Dec 1 '18 at 14:39


















  • Is it propositional logic? How is $models$ defined?
    – Berci
    Dec 1 '18 at 14:28










  • @Berci I added my definition.
    – heaig
    Dec 1 '18 at 14:39






  • 1




    Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
    – Mauro ALLEGRANZA
    Dec 1 '18 at 14:39
















Is it propositional logic? How is $models$ defined?
– Berci
Dec 1 '18 at 14:28




Is it propositional logic? How is $models$ defined?
– Berci
Dec 1 '18 at 14:28












@Berci I added my definition.
– heaig
Dec 1 '18 at 14:39




@Berci I added my definition.
– heaig
Dec 1 '18 at 14:39




1




1




Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39




Consider the simple case of $k=2$. If $(H_1 land H_2) to F$ is taut, this means that the formula is true for every truth assignment. Thus, what happens with a truth assignment that satisfies $(H_1 land H_2)$ ?
– Mauro ALLEGRANZA
Dec 1 '18 at 14:39










2 Answers
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Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.



So we have:



$$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$



Introduce preconditions on both sides:



$${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$



Simplify the implication:



$${H_1 wedge cdots wedge H_k} models F$$






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    0














    To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      0














      Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.



      So we have:



      $$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$



      Introduce preconditions on both sides:



      $${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$



      Simplify the implication:



      $${H_1 wedge cdots wedge H_k} models F$$






      share|cite|improve this answer


























        0














        Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.



        So we have:



        $$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$



        Introduce preconditions on both sides:



        $${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$



        Simplify the implication:



        $${H_1 wedge cdots wedge H_k} models F$$






        share|cite|improve this answer
























          0












          0








          0






          Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.



          So we have:



          $$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$



          Introduce preconditions on both sides:



          $${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$



          Simplify the implication:



          $${H_1 wedge cdots wedge H_k} models F$$






          share|cite|improve this answer












          Note that you can write "$X$ is a tautology" as "$emptysetmodels X$", with the empty set of preconditions.



          So we have:



          $$emptysetmodels (H_1 wedge cdots wedge H_k) to F$$



          Introduce preconditions on both sides:



          $${H_1 wedge cdots wedge H_k} models (H_1 wedge cdots wedge H_k) wedge ((H_1 wedge cdots wedge H_k) to F)$$



          Simplify the implication:



          $${H_1 wedge cdots wedge H_k} models F$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 14:48









          orlp

          7,3411230




          7,3411230























              0














              To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.






              share|cite|improve this answer


























                0














                To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.






                share|cite|improve this answer
























                  0












                  0








                  0






                  To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.






                  share|cite|improve this answer












                  To say that ${H_1,dots,H_k}models F$ is to say that there is no interpretation making all the $H_i$'s true and making $F$ false. To say that $(H_1landdotsland H_k)to F$ is tautology means, thanks to the truth table for implication, that there is no interpretation making $H_1landdotsland H_k$ true and making $F$ false. By the truth table for conjunction, these are the same.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 15:05









                  Andreas Blass

                  49.3k351106




                  49.3k351106






























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