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Definitional equality of two propositions about propositional equality


Why values can not be replaced with their extensionally equal values in an intensional system?Intro to Martin-Löf type theoryNon-termination of types in Martin-Löf's Type:Type?Why is `map insertionsort` not to equal to`map mergesort`?Indexing a dependent type on a value?How Types avoids Russel's Paradox?What fragment of Martin-Löf dependent type theory can be expressed using generic types in Java?Extensional constructs in minimal extensional type theory without eta equalityMeaning of the notation Typ := TVar | (Typ → Typ)Dependent Type Theory Implementation of a GraphPropositional extentionality in the lean theorem prover?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


Martin Hofmann states in Extensional Concepts in Intensional Type Theory (§1.1 p.3) that:




It is important that definition equality is pervasive so if M and N are definitionally equal then P(M) is definitionally equal to P(N) whatever P is. In particular the proposition stating that M is propositionally equal to N is definitionally equal to the proposition that M is propositionally equal to itself.




What I don't get is how a proposition about equality of M and N propositionally can be definitionally equal to the proposition about propositional equality of M and M ?










share|cite|improve this question











$endgroup$




















    4












    $begingroup$


    Martin Hofmann states in Extensional Concepts in Intensional Type Theory (§1.1 p.3) that:




    It is important that definition equality is pervasive so if M and N are definitionally equal then P(M) is definitionally equal to P(N) whatever P is. In particular the proposition stating that M is propositionally equal to N is definitionally equal to the proposition that M is propositionally equal to itself.




    What I don't get is how a proposition about equality of M and N propositionally can be definitionally equal to the proposition about propositional equality of M and M ?










    share|cite|improve this question











    $endgroup$
















      4












      4








      4





      $begingroup$


      Martin Hofmann states in Extensional Concepts in Intensional Type Theory (§1.1 p.3) that:




      It is important that definition equality is pervasive so if M and N are definitionally equal then P(M) is definitionally equal to P(N) whatever P is. In particular the proposition stating that M is propositionally equal to N is definitionally equal to the proposition that M is propositionally equal to itself.




      What I don't get is how a proposition about equality of M and N propositionally can be definitionally equal to the proposition about propositional equality of M and M ?










      share|cite|improve this question











      $endgroup$




      Martin Hofmann states in Extensional Concepts in Intensional Type Theory (§1.1 p.3) that:




      It is important that definition equality is pervasive so if M and N are definitionally equal then P(M) is definitionally equal to P(N) whatever P is. In particular the proposition stating that M is propositionally equal to N is definitionally equal to the proposition that M is propositionally equal to itself.




      What I don't get is how a proposition about equality of M and N propositionally can be definitionally equal to the proposition about propositional equality of M and M ?







      type-theory dependent-types equality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 6 at 19:03









      Gilles

      34.6k7 gold badges99 silver badges166 bronze badges




      34.6k7 gold badges99 silver badges166 bronze badges










      asked Aug 6 at 15:26









      al palal pal

      925 bronze badges




      925 bronze badges























          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Recall that (a few paragraphs above)




          two objects are definitionally equal if after certain computation steps they evaluate
          to identical results.




          Assume throughout this post that $M$ and $N$ are definitionally equal. This means that there is a series of computation steps $M_0 leftrightarrow M_1 leftrightarrow ldots leftrightarrow M_n$ where I use $leftrightarrow$ do denote a computation step, and $M_0$ is the term $M$ and $M_n$ is the term $N$.



          Any computation step can be applied in any context. This is part of the definition of computation rules for calculi used in logic (as opposed to calculi used to model programming languages with side effects). So if you stick $M_0$ in some context $P$, there is a series of computation steps $P(M_0) leftrightarrow P(M_1) leftrightarrow ldots leftrightarrow P(M_n)$. This means that $P(M_0)$ and $P(M_n)$ are definitionally equal, i.e. $P(M)$ and $P(N)$ are definitionally equal.



          Apply this to a property $P(X)$ that expresses “$M$ is propositionally equal to $X$”, no matter how this is expressed in a particular calculus. $P(M)$ is definitionally equal to $P(N)$, i.e. a property that expresses “$M$ is propositionally equal to $M$” is definitionally equal to “$M$ is propositionally equal to $N$”.



          Strictly speaking, there isn't necessarily a single proposition stating that $M$ is propositionally equal to $N$ — there may be ways to state this that are not definitionally equal (it depends on the exact calculus. If a calculus has multiple ways to express this, there can be two (or more) non-definitionally-equal propositions stating that $M$ is equal to itself, and then there are also multiple non-definitionally-equal propositions stating that $M$ is equal to $N$. But for each proposition stating that $M$ is equal to itself, that proposition is definitionally equal to a proposition stating that $M$ is equal to $N$, by applying the computation steps that rewrite $M$ to $N$.



          Reasoning about equality can be difficult to follow. We tend to have an intuition that equality is just equality, dammit. It's a bit difficult to internalize that there are different concepts of equality, and worse, they coexist in the same theory. I think it helps if you use words other than “equality”. If two terms are written exactly in the same way (as strings or at least as abstract syntax trees), they're identical. As long as you're not reasoning about variable binding, you can extend that to terms that are identical except for variable names, i.e. terms that are alpha-equivalent. Beyond that, don't use the word “equal”. If there's a series of computation steps that leads from $M$ to $N$, say that $M$ and $N$ are computationally equivalent or rewritable to each other. They're not necessarily “equal”, but they're equivalent for a certain equivalence relation. Then it's obvious and not particularly confusing that there can be coarser equivalences between terms. For example, there's an equivalence relation between terms defined by “if you replace $M$ by $N$ in a true proposition, you get a proposition that's still true”. This is a form of observational or extensional equivalence of propositions. It can be called “observational equality” or “propositional equality”, but to form an intuition, don't call it that. And then it's not so shocking that if $N$ can be rewritten to $M$, then any proposition stating that $M$ is whatever-equivalent to $N$ can be rewritten to a proposition that states that $M$ is equivalent to itself.






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
            $endgroup$
            – al pal
            Aug 6 at 22:04










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            Aug 6 at 22:07













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          1 Answer
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          active

          oldest

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          5












          $begingroup$

          Recall that (a few paragraphs above)




          two objects are definitionally equal if after certain computation steps they evaluate
          to identical results.




          Assume throughout this post that $M$ and $N$ are definitionally equal. This means that there is a series of computation steps $M_0 leftrightarrow M_1 leftrightarrow ldots leftrightarrow M_n$ where I use $leftrightarrow$ do denote a computation step, and $M_0$ is the term $M$ and $M_n$ is the term $N$.



          Any computation step can be applied in any context. This is part of the definition of computation rules for calculi used in logic (as opposed to calculi used to model programming languages with side effects). So if you stick $M_0$ in some context $P$, there is a series of computation steps $P(M_0) leftrightarrow P(M_1) leftrightarrow ldots leftrightarrow P(M_n)$. This means that $P(M_0)$ and $P(M_n)$ are definitionally equal, i.e. $P(M)$ and $P(N)$ are definitionally equal.



          Apply this to a property $P(X)$ that expresses “$M$ is propositionally equal to $X$”, no matter how this is expressed in a particular calculus. $P(M)$ is definitionally equal to $P(N)$, i.e. a property that expresses “$M$ is propositionally equal to $M$” is definitionally equal to “$M$ is propositionally equal to $N$”.



          Strictly speaking, there isn't necessarily a single proposition stating that $M$ is propositionally equal to $N$ — there may be ways to state this that are not definitionally equal (it depends on the exact calculus. If a calculus has multiple ways to express this, there can be two (or more) non-definitionally-equal propositions stating that $M$ is equal to itself, and then there are also multiple non-definitionally-equal propositions stating that $M$ is equal to $N$. But for each proposition stating that $M$ is equal to itself, that proposition is definitionally equal to a proposition stating that $M$ is equal to $N$, by applying the computation steps that rewrite $M$ to $N$.



          Reasoning about equality can be difficult to follow. We tend to have an intuition that equality is just equality, dammit. It's a bit difficult to internalize that there are different concepts of equality, and worse, they coexist in the same theory. I think it helps if you use words other than “equality”. If two terms are written exactly in the same way (as strings or at least as abstract syntax trees), they're identical. As long as you're not reasoning about variable binding, you can extend that to terms that are identical except for variable names, i.e. terms that are alpha-equivalent. Beyond that, don't use the word “equal”. If there's a series of computation steps that leads from $M$ to $N$, say that $M$ and $N$ are computationally equivalent or rewritable to each other. They're not necessarily “equal”, but they're equivalent for a certain equivalence relation. Then it's obvious and not particularly confusing that there can be coarser equivalences between terms. For example, there's an equivalence relation between terms defined by “if you replace $M$ by $N$ in a true proposition, you get a proposition that's still true”. This is a form of observational or extensional equivalence of propositions. It can be called “observational equality” or “propositional equality”, but to form an intuition, don't call it that. And then it's not so shocking that if $N$ can be rewritten to $M$, then any proposition stating that $M$ is whatever-equivalent to $N$ can be rewritten to a proposition that states that $M$ is equivalent to itself.






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
            $endgroup$
            – al pal
            Aug 6 at 22:04










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            Aug 6 at 22:07















          5












          $begingroup$

          Recall that (a few paragraphs above)




          two objects are definitionally equal if after certain computation steps they evaluate
          to identical results.




          Assume throughout this post that $M$ and $N$ are definitionally equal. This means that there is a series of computation steps $M_0 leftrightarrow M_1 leftrightarrow ldots leftrightarrow M_n$ where I use $leftrightarrow$ do denote a computation step, and $M_0$ is the term $M$ and $M_n$ is the term $N$.



          Any computation step can be applied in any context. This is part of the definition of computation rules for calculi used in logic (as opposed to calculi used to model programming languages with side effects). So if you stick $M_0$ in some context $P$, there is a series of computation steps $P(M_0) leftrightarrow P(M_1) leftrightarrow ldots leftrightarrow P(M_n)$. This means that $P(M_0)$ and $P(M_n)$ are definitionally equal, i.e. $P(M)$ and $P(N)$ are definitionally equal.



          Apply this to a property $P(X)$ that expresses “$M$ is propositionally equal to $X$”, no matter how this is expressed in a particular calculus. $P(M)$ is definitionally equal to $P(N)$, i.e. a property that expresses “$M$ is propositionally equal to $M$” is definitionally equal to “$M$ is propositionally equal to $N$”.



          Strictly speaking, there isn't necessarily a single proposition stating that $M$ is propositionally equal to $N$ — there may be ways to state this that are not definitionally equal (it depends on the exact calculus. If a calculus has multiple ways to express this, there can be two (or more) non-definitionally-equal propositions stating that $M$ is equal to itself, and then there are also multiple non-definitionally-equal propositions stating that $M$ is equal to $N$. But for each proposition stating that $M$ is equal to itself, that proposition is definitionally equal to a proposition stating that $M$ is equal to $N$, by applying the computation steps that rewrite $M$ to $N$.



          Reasoning about equality can be difficult to follow. We tend to have an intuition that equality is just equality, dammit. It's a bit difficult to internalize that there are different concepts of equality, and worse, they coexist in the same theory. I think it helps if you use words other than “equality”. If two terms are written exactly in the same way (as strings or at least as abstract syntax trees), they're identical. As long as you're not reasoning about variable binding, you can extend that to terms that are identical except for variable names, i.e. terms that are alpha-equivalent. Beyond that, don't use the word “equal”. If there's a series of computation steps that leads from $M$ to $N$, say that $M$ and $N$ are computationally equivalent or rewritable to each other. They're not necessarily “equal”, but they're equivalent for a certain equivalence relation. Then it's obvious and not particularly confusing that there can be coarser equivalences between terms. For example, there's an equivalence relation between terms defined by “if you replace $M$ by $N$ in a true proposition, you get a proposition that's still true”. This is a form of observational or extensional equivalence of propositions. It can be called “observational equality” or “propositional equality”, but to form an intuition, don't call it that. And then it's not so shocking that if $N$ can be rewritten to $M$, then any proposition stating that $M$ is whatever-equivalent to $N$ can be rewritten to a proposition that states that $M$ is equivalent to itself.






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
            $endgroup$
            – al pal
            Aug 6 at 22:04










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            Aug 6 at 22:07













          5












          5








          5





          $begingroup$

          Recall that (a few paragraphs above)




          two objects are definitionally equal if after certain computation steps they evaluate
          to identical results.




          Assume throughout this post that $M$ and $N$ are definitionally equal. This means that there is a series of computation steps $M_0 leftrightarrow M_1 leftrightarrow ldots leftrightarrow M_n$ where I use $leftrightarrow$ do denote a computation step, and $M_0$ is the term $M$ and $M_n$ is the term $N$.



          Any computation step can be applied in any context. This is part of the definition of computation rules for calculi used in logic (as opposed to calculi used to model programming languages with side effects). So if you stick $M_0$ in some context $P$, there is a series of computation steps $P(M_0) leftrightarrow P(M_1) leftrightarrow ldots leftrightarrow P(M_n)$. This means that $P(M_0)$ and $P(M_n)$ are definitionally equal, i.e. $P(M)$ and $P(N)$ are definitionally equal.



          Apply this to a property $P(X)$ that expresses “$M$ is propositionally equal to $X$”, no matter how this is expressed in a particular calculus. $P(M)$ is definitionally equal to $P(N)$, i.e. a property that expresses “$M$ is propositionally equal to $M$” is definitionally equal to “$M$ is propositionally equal to $N$”.



          Strictly speaking, there isn't necessarily a single proposition stating that $M$ is propositionally equal to $N$ — there may be ways to state this that are not definitionally equal (it depends on the exact calculus. If a calculus has multiple ways to express this, there can be two (or more) non-definitionally-equal propositions stating that $M$ is equal to itself, and then there are also multiple non-definitionally-equal propositions stating that $M$ is equal to $N$. But for each proposition stating that $M$ is equal to itself, that proposition is definitionally equal to a proposition stating that $M$ is equal to $N$, by applying the computation steps that rewrite $M$ to $N$.



          Reasoning about equality can be difficult to follow. We tend to have an intuition that equality is just equality, dammit. It's a bit difficult to internalize that there are different concepts of equality, and worse, they coexist in the same theory. I think it helps if you use words other than “equality”. If two terms are written exactly in the same way (as strings or at least as abstract syntax trees), they're identical. As long as you're not reasoning about variable binding, you can extend that to terms that are identical except for variable names, i.e. terms that are alpha-equivalent. Beyond that, don't use the word “equal”. If there's a series of computation steps that leads from $M$ to $N$, say that $M$ and $N$ are computationally equivalent or rewritable to each other. They're not necessarily “equal”, but they're equivalent for a certain equivalence relation. Then it's obvious and not particularly confusing that there can be coarser equivalences between terms. For example, there's an equivalence relation between terms defined by “if you replace $M$ by $N$ in a true proposition, you get a proposition that's still true”. This is a form of observational or extensional equivalence of propositions. It can be called “observational equality” or “propositional equality”, but to form an intuition, don't call it that. And then it's not so shocking that if $N$ can be rewritten to $M$, then any proposition stating that $M$ is whatever-equivalent to $N$ can be rewritten to a proposition that states that $M$ is equivalent to itself.






          share|cite|improve this answer











          $endgroup$



          Recall that (a few paragraphs above)




          two objects are definitionally equal if after certain computation steps they evaluate
          to identical results.




          Assume throughout this post that $M$ and $N$ are definitionally equal. This means that there is a series of computation steps $M_0 leftrightarrow M_1 leftrightarrow ldots leftrightarrow M_n$ where I use $leftrightarrow$ do denote a computation step, and $M_0$ is the term $M$ and $M_n$ is the term $N$.



          Any computation step can be applied in any context. This is part of the definition of computation rules for calculi used in logic (as opposed to calculi used to model programming languages with side effects). So if you stick $M_0$ in some context $P$, there is a series of computation steps $P(M_0) leftrightarrow P(M_1) leftrightarrow ldots leftrightarrow P(M_n)$. This means that $P(M_0)$ and $P(M_n)$ are definitionally equal, i.e. $P(M)$ and $P(N)$ are definitionally equal.



          Apply this to a property $P(X)$ that expresses “$M$ is propositionally equal to $X$”, no matter how this is expressed in a particular calculus. $P(M)$ is definitionally equal to $P(N)$, i.e. a property that expresses “$M$ is propositionally equal to $M$” is definitionally equal to “$M$ is propositionally equal to $N$”.



          Strictly speaking, there isn't necessarily a single proposition stating that $M$ is propositionally equal to $N$ — there may be ways to state this that are not definitionally equal (it depends on the exact calculus. If a calculus has multiple ways to express this, there can be two (or more) non-definitionally-equal propositions stating that $M$ is equal to itself, and then there are also multiple non-definitionally-equal propositions stating that $M$ is equal to $N$. But for each proposition stating that $M$ is equal to itself, that proposition is definitionally equal to a proposition stating that $M$ is equal to $N$, by applying the computation steps that rewrite $M$ to $N$.



          Reasoning about equality can be difficult to follow. We tend to have an intuition that equality is just equality, dammit. It's a bit difficult to internalize that there are different concepts of equality, and worse, they coexist in the same theory. I think it helps if you use words other than “equality”. If two terms are written exactly in the same way (as strings or at least as abstract syntax trees), they're identical. As long as you're not reasoning about variable binding, you can extend that to terms that are identical except for variable names, i.e. terms that are alpha-equivalent. Beyond that, don't use the word “equal”. If there's a series of computation steps that leads from $M$ to $N$, say that $M$ and $N$ are computationally equivalent or rewritable to each other. They're not necessarily “equal”, but they're equivalent for a certain equivalence relation. Then it's obvious and not particularly confusing that there can be coarser equivalences between terms. For example, there's an equivalence relation between terms defined by “if you replace $M$ by $N$ in a true proposition, you get a proposition that's still true”. This is a form of observational or extensional equivalence of propositions. It can be called “observational equality” or “propositional equality”, but to form an intuition, don't call it that. And then it's not so shocking that if $N$ can be rewritten to $M$, then any proposition stating that $M$ is whatever-equivalent to $N$ can be rewritten to a proposition that states that $M$ is equivalent to itself.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 6 at 22:07

























          answered Aug 6 at 19:02









          GillesGilles

          34.6k7 gold badges99 silver badges166 bronze badges




          34.6k7 gold badges99 silver badges166 bronze badges














          • $begingroup$
            Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
            $endgroup$
            – al pal
            Aug 6 at 22:04










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            Aug 6 at 22:07
















          • $begingroup$
            Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
            $endgroup$
            – al pal
            Aug 6 at 22:04










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            Aug 6 at 22:07















          $begingroup$
          Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
          $endgroup$
          – al pal
          Aug 6 at 22:04




          $begingroup$
          Thanks for the answer. So in your third paragraph when you talk about P(X), P(M) and P(N), the presumption is that M and N are already definitionally equal, did I get it right?
          $endgroup$
          – al pal
          Aug 6 at 22:04












          $begingroup$
          @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
          $endgroup$
          – Gilles
          Aug 6 at 22:07




          $begingroup$
          @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
          $endgroup$
          – Gilles
          Aug 6 at 22:07

















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