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authorGravatar herbelin <herbelin@85f007b7-540e-0410-9357-904b9bb8a0f7>2003-04-07 20:25:59 +0000
committerGravatar herbelin <herbelin@85f007b7-540e-0410-9357-904b9bb8a0f7>2003-04-07 20:25:59 +0000
commit7590031325b144f6864d2194a700a92145617c13 (patch)
treeed9b81f7d5cf07706d137a87c6e7e1c609f57d8e /doc
parent3cf2de31eccc27952075d25edf57524525509b36 (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.tex20
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.