X-Git-Url: http://matita.cs.unibo.it/gitweb/?a=blobdiff_plain;f=helm%2Fsoftware%2Fmatita%2Flibrary%2Fdemo%2Fpower_derivative.ma;h=76fa66115daa741db40ed4aa04502513da6b9b54;hb=c00f22f7afa508881c8d116928e1c460600ba0ac;hp=4f1b44000889a1df1b765d98a29977b8a3b2a432;hpb=5649890273cf8e660bba744e84ce5fee1e5efe69;p=helm.git

diff --git a/helm/software/matita/library/demo/power_derivative.ma b/helm/software/matita/library/demo/power_derivative.ma
index 4f1b44000..76fa66115 100644
--- a/helm/software/matita/library/demo/power_derivative.ma
+++ b/helm/software/matita/library/demo/power_derivative.ma
@@ -12,8 +12,6 @@
 (*                                                                        *)
 (**************************************************************************)
 
-set "baseuri" "cic:/matita/demo/power_derivative".
-
 include "nat/plus.ma".
 include "nat/orders.ma".
 include "nat/compare.ma".
@@ -256,7 +254,6 @@ theorem derivative_power: ∀n:nat. D[x \sup n] = n·x \sup (pred n).
  elim n 0.
  case O.
    the thesis becomes (D[x \sup 0] = 0·x \sup (pred 0)).
-   by _
   done.
  case S (m:nat).
   by induction hypothesis we know
@@ -265,24 +262,24 @@ theorem derivative_power: ∀n:nat. D[x \sup n] = n·x \sup (pred n).
    (D[x \sup (1+m)] = (1+m) · x \sup m).
   we need to prove
    (m · (x \sup (1+ pred m)) = m · x \sup m) (Ppred).
-   by _ we proved (0 < m ∨ 0=m) (cases).
+   we proved (0 < m ∨ 0=m) (cases).
    we proceed by induction on cases
    to prove (m · (x \sup (1+ pred m)) = m · x \sup m).
     case left.
       suppose (0 < m) (m_pos).
-      by (S_pred m m_pos) we proved (m = 1 + pred m) (H1).
-      by _
+      using (S_pred ? m_pos) we proved (m = 1 + pred m) (H1).
      done.
     case right.
-      suppose (0=m) (m_zero). by _ done.
+      suppose (0=m) (m_zero). 
+    done.
   conclude
      (D[x \sup (1+m)])
    = (D[x · x \sup m]).
    = (D[x] · x \sup m + x · D[x \sup m]).
-   = (x \sup m + x · (m · x \sup (pred m))).
+   = (x \sup m + x · (m · x \sup (pred m))) timeout=30.
    = (x \sup m + m · (x \sup (1 + pred m))).
    = (x \sup m + m · x \sup m).
-   = ((1+m) · x \sup m) by Fmult_one_f Fmult_commutative Fmult_Fplus_distr costante_sum
+   = ((1+m) · x \sup m) timeout=30 by Fmult_one_f, Fmult_commutative, Fmult_Fplus_distr, costante_sum
   done.
 qed.
 
@@ -310,7 +307,6 @@ theorem derivative_power': ∀n:nat. D[x \sup (1+n)] = (1+n) · x \sup n.
  (D[x \sup (1+n)] = (1+n) · x \sup n).*) elim n 0.
  case O.
    the thesis becomes (D[x \sup 1] = 1 · x \sup 0).
-   by _
   done.
  case S (m:nat).
   by induction hypothesis we know
@@ -323,75 +319,7 @@ 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.
+   = ((2+m) · x \sup (1+m)) timeout=30 by Fmult_one_f, Fmult_commutative,
+       Fmult_Fplus_distr, assoc_plus, plus_n_SO, costante_sum
+  done.
 qed.