Definitional equality of two propositions about propositional equalityIntro to Martin-Löf type...

If someone else uploads my GPL'd code to Github without my permission, is that a copyright violation?

What was the role of Commodore-West Germany?

Why does capacitance not depend on the material of the plates?

Getting an entry level IT position later in life

Does a humanoid possessed by a ghost register as undead to a paladin's Divine Sense?

Does the length of a password for Wi-Fi affect speed?

Launch capabilities of GSLV Mark III

Definitional equality of two propositions about propositional equality

Make a living as a math programming freelancer?

How to check a file was encrypted (really & correctly)

Does a 4 bladed prop have almost twice the thrust of a 2 bladed prop?

Write The Shortest Program To Check If A Binary Tree Is Balanced

Nested Unlocked Packages requires Installation of Base Packages?

Why private jets such as GulfStream ones fly higher than other civil jets?

Premier League simulation

London underground zone 1-2 train ticket

Can attackers change the public key of certificate during the SSL handshake

Can a Hogwarts student refuse the Sorting Hat's decision?

How and where to get you research work assessed for PhD?

What prevents ads from reading my password as I type it?

Can I enter a rental property without giving notice if I'm afraid a tenant may be hurt?

How do I get the =LEFT function in excel, to also take the number zero as the first number?

Purchased new computer from DELL with pre-installed Ubuntu. Won't boot. Should assume its an error from DELL?

Should I take out a personal loan to pay off credit card debt?



Definitional equality of two propositions about propositional equality


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;
}







3












$begingroup$


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











$endgroup$





















    3












    $begingroup$


    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











    $endgroup$

















      3












      3








      3





      $begingroup$


      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 5 hours ago









      Gilles

      34.4k7 gold badges98 silver badges166 bronze badges




      34.4k7 gold badges98 silver badges166 bronze badges










      asked 9 hours ago









      al palal pal

      403 bronze badges




      403 bronze badges

























          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          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
            2 hours ago










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            2 hours ago














          Your Answer








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "419"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f112501%2fdefinitional-equality-of-two-propositions-about-propositional-equality%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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
            2 hours ago










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            2 hours ago
















          4












          $begingroup$

          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
            2 hours ago










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            2 hours ago














          4












          4








          4





          $begingroup$

          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 2 hours ago

























          answered 5 hours ago









          GillesGilles

          34.4k7 gold badges98 silver badges166 bronze badges




          34.4k7 gold badges98 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
            2 hours ago










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            2 hours ago


















          • $begingroup$
            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
            2 hours ago










          • $begingroup$
            @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
            $endgroup$
            – Gilles
            2 hours ago
















          $begingroup$
          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
          2 hours ago




          $begingroup$
          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
          2 hours ago












          $begingroup$
          @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
          $endgroup$
          – Gilles
          2 hours ago




          $begingroup$
          @alpal Yes, it's all a continuation of “If M and N are computationally equal” near the beginning.
          $endgroup$
          – Gilles
          2 hours ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Computer Science Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f112501%2fdefinitional-equality-of-two-propositions-about-propositional-equality%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Taj Mahal Inhaltsverzeichnis Aufbau | Geschichte | 350-Jahr-Feier | Heutige Bedeutung | Siehe auch |...

          Baia Sprie Cuprins Etimologie | Istorie | Demografie | Politică și administrație | Arii naturale...

          Nicolae Petrescu-Găină Cuprins Biografie | Opera | In memoriam | Varia | Controverse, incertitudini...