From: Claudio Sacerdoti Coen <claudio.sacerdoticoen@unibo.it>
Date: Thu, 20 Mar 2008 20:39:28 +0000 (+0000)
Subject: Script fixed (it did not compile due to a mistake before committing).
X-Git-Tag: make_still_working~5510
X-Git-Url: http://matita.cs.unibo.it/gitweb/?a=commitdiff_plain;h=9e7db2554d0fd904966c0e6988f1c8763dd15a0b;p=helm.git

Script fixed (it did not compile due to a mistake before committing).
---

diff --git a/helm/software/matita/library/demo/power_derivative.ma b/helm/software/matita/library/demo/power_derivative.ma
index 4f1b44000..fd58c9564 100644
--- a/helm/software/matita/library/demo/power_derivative.ma
+++ b/helm/software/matita/library/demo/power_derivative.ma
@@ -323,75 +323,7 @@ theorem derivative_power': ∀n:nat. D[x \sup (1+n)] = (1+n) · x \sup n.
    = (D[x] · x \sup (1+m) + x · D[x \sup (1+m)]).
    = (x \sup (1+m) + x · (costante (1+m) · x \sup m)).
    = (x \sup (1+m) + costante (1+m) · x \sup (1+m)).
-
-
-
-  conclude (x \sup (1+m) + costante (1+m) · x \sup (1+m))
-  = (costante 1 · x \sup (1+m) + costante (1+m) ·(x) \sup (1+m)) proof.
-   by (Fmult_one_f ((x)\sup(1+m)))
-   we proved (costante 1·(x)\sup(1+m)=(x)\sup(1+m)) (previous).
-   by (eq_OF_eq ? ? (λfoo:(R→R).foo+costante (1+m)·(x)\sup(1+m)) (costante 1
-                                                                                                                                                                                                                                       ·(x)\sup(1
-                                                                                                                                                                                                                                                    +m)) ((x)\sup(1
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             +m)) previous)
-   done.
-  = ((x)\sup(1+m)·costante 1+costante (1+m)·(x)\sup(1+m)) proof.
-   by (Fmult_commutative (costante 1) ((x)\sup(1+m)))
-   we proved (costante 1·(x)\sup(1+m)=(x)\sup(1+m)·costante 1) (previous).
-   by (eq_f ? ? (λfoo:(R→R).foo+costante (1+m)·(x)\sup(1+m)) (costante 1
-                                                                                                                                                                                                       ·(x)\sup(1+m)) ((x)\sup(1
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         +m)
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              ·costante
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                1) previous)
-   done.
-  = ((x)\sup(1+m)·costante 1+(x)\sup(1+m)·costante (1+m)) proof.
-   by (Fmult_commutative ((x)\sup(1+m)) (costante (1+m)))
-   we proved ((x)\sup(1+m)·costante (1+m)=costante (1+m)·(x)\sup(1+m))
-    
-   (previous)
-   .
-   by (eq_OF_eq ? ? (λfoo:(R→R).(x)\sup(1+m)·costante 1+foo) ((x)\sup(1+m)
-                                                                                                                                                                                                                               ·costante
-                                                                                                                                                                                                                                 (1+m)) (costante
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        (1
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         +m)
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        ·(x)\sup(1
-                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     +m)) previous)
-   done.
-  = ((x)\sup(1+m)·(costante 1+costante (1+m))) proof.
-   by (Fmult_Fplus_distr ((x)\sup(1+m)) (costante 1) (costante (1+m)))
-   we proved
-   ((x)\sup(1+m)·(costante 1+costante (1+m))
-    =(x)\sup(1+m)·costante 1+(x)\sup(1+m)·costante (1+m))
-    
-   (previous)
-   .
-   by (sym_eq ? ((x)\sup(1+m)·(costante 1+costante (1+m))) ((x)\sup(1+m)
-                                                                                                                                                                            ·costante 1
-                                                                                                                                                                            +(x)\sup(1+m)
-                                                                                                                                                                             ·costante (1+m)) previous)
-   done.
-  = ((costante 1+costante (1+m))·(x)\sup(1+m)) 
-  exact (Fmult_commutative ((x)\sup(1+m)) (costante 1+costante (1+m))).
-  = (costante (1+(1+m))·(x)\sup(1+m)) proof.
-   by (costante_sum 1 (1+m))
-   we proved (costante 1+costante (1+m)=costante (1+(1+m))) (previous).
-   by (eq_f ? ? (λfoo:(R→R).foo·(x)\sup(1+m)) (costante 1+costante (1+m)) (costante
-                                                                                                                                                                                                                                                                                                                              (1
-                                                                                                                                                                                                                                                                                                                               +(1
-                                                                                                                                                                                                                                                                                                                                 +m))) previous)
-   done.
-  = (costante (1+1+m)·(x)\sup(1+m)) proof.
-   by (assoc_plus 1 1 m)
-   we proved (1+1+m=1+(1+m)) (previous).
-   by (eq_OF_eq ? ? (λfoo:nat.costante foo·(x)\sup(1+m)) ? ? previous)
-   done.
-  = (costante (2+m)·(x)\sup(1+m)) proof done.
-   by (plus_n_SO 1)
-   we proved (2=1+1) (previous).
-   by (eq_OF_eq ? ? (λfoo:nat.costante (foo+m)·(x)\sup(1+m)) ? ? previous)
-   done.
-  
- 
-(* end auto($Revision: 8206 $) proof: TIME=0.06 SIZE=100 DEPTH=100 *)  done.
-qed.
+   = ((2+m) · x \sup (1+m)) by Fmult_one_f Fmult_commutative
+     Fmult_Fplus_distr assoc_plus plus_n_SO costante_sum
+  done.
+qed.
\ No newline at end of file