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author | herbelin <herbelin@85f007b7-540e-0410-9357-904b9bb8a0f7> | 2003-04-07 20:25:59 +0000 |
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committer | herbelin <herbelin@85f007b7-540e-0410-9357-904b9bb8a0f7> | 2003-04-07 20:25:59 +0000 |
commit | 7590031325b144f6864d2194a700a92145617c13 (patch) | |
tree | ed9b81f7d5cf07706d137a87c6e7e1c609f57d8e /doc | |
parent | 3cf2de31eccc27952075d25edf57524525509b36 (diff) |
Divers
git-svn-id: svn+ssh://scm.gforge.inria.fr/svn/coq/trunk@8333 85f007b7-540e-0410-9357-904b9bb8a0f7
Diffstat (limited to 'doc')
-rw-r--r-- | doc/faq.tex | 20 |
1 files changed, 14 insertions, 6 deletions
diff --git a/doc/faq.tex b/doc/faq.tex index fb73ee5a3..c69c587b7 100644 --- a/doc/faq.tex +++ b/doc/faq.tex @@ -204,15 +204,15 @@ Axiom UIP_refl: (A:Set)(x:A)(p:x=x)p==(refl_equal A x). Axiom $K$ is also equivalent to \begin{coq_example*} -Axiom eq_rec_eq : (a:A)(P:A->Set)(p:(P p))(h:a==a) p=(eq_rect A a P p a h). +Axiom eq_rec_eq : (A:Set)(a:A)(P:A->Set)(p:(P p))(h:a==a) p=(eq_rect A a P p a h). \end{coq_example*} It is also equivalent to the injectivity of dependent equality (dependent equality is itself equivalent to equality of dependent pairs). \begin{coq_example*} -Inductive eq_dep [p:U;x:(P p)] : (q:U)(P q)->Prop := - eq_dep_intro : (eq_dep p x p x). -Axiom eq_dep_eq : (a:A)(P:A->Set)(p1,p2:(P a))(eq_dep a p1 a p2)->p1=p2. +Inductive eq_dep [U:Set;P:U->Set;p:U;x:(P p)] : (q:U)(P q)->Prop := + eq_dep_intro : (eq_dep U P p x p x). +Axiom eq_dep_eq : (U:Set)(u:U)(P:U->Set)(p1,p2:(P u))(eq_dep U P u p1 u p2)->p1=p2. \end{coq_example*} All of these statements can be found in file \vfile{\LogicEqdep}{Eqdep}. @@ -262,7 +262,7 @@ Exact [x]x. Rewrite H in Hdiscr. Elim Hdiscr. Assert H1:=(snd (exist nat [z](le z (S n)) (S n) (le_n (S n)))). -... +(* ... *) \end{coq_example*} \section{Axioms} @@ -332,6 +332,10 @@ e.g. Burali-Forti paradox \cite{Gir70,Coq85}. \question{Why Injection does not work on impredicative types?} E.g. in this case: +\begin{coq_eval} +Reset Initial. +\end{coq_eval} + \begin{coq_example*} Inductive I : Set := intro : (k:Set)k -> I. Lemma eq_jdef : (x,y:nat) (intro ? x)=(intro ? y) -> x=y. @@ -613,6 +617,10 @@ Lemma eq_comb \answer You need to derive the dependent elimination scheme for DefProp by hand using {\coqtt Scheme}. +\begin{coq_eval} +Abort. +\end{coq_eval} + \begin{coq_example*} Scheme DefProp_elim := Induction for DefProp Sort Prop. @@ -647,7 +655,7 @@ Lemma eq_pack \answer \begin{coq_eval} -Reset eq_pack. +Abort. \end{coq_eval} \begin{coq_example*} Scheme natProp_elim := Induction for natProp Sort Prop. |