simplify;
apply f_eq_extensional;
intro;
- auto.
+ autobatch.
qed.
lemma Fmult_zero_f: ∀f:R→R.0·f=0.
simplify;
apply f_eq_extensional;
intro;
- auto.
+ autobatch.
qed.
lemma Fmult_commutative: symmetric ? Fmult.
unfold Fmult;
apply f_eq_extensional;
intros;
- auto.
+ autobatch.
qed.
lemma Fmult_associative: associative ? Fmult.
unfold Fmult;
apply f_eq_extensional;
intros;
- auto.
+ autobatch.
qed.
lemma Fmult_Fplus_distr: distributive ? Fmult Fplus.
apply f_eq_extensional;
intros;
simplify;
- auto.
+ autobatch.
qed.
lemma monomio_product:
[ simplify;
apply f_eq_extensional;
intro;
- auto
+ autobatch
| simplify;
apply f_eq_extensional;
intro;
cut (x\sup (n1+m) = x \sup n1 · x \sup m);
[ rewrite > Hcut;
- auto
+ autobatch
| change in ⊢ (? ? % ?) with ((λx:R.x\sup(n1+m)) x);
rewrite > H;
reflexivity
intros;
elim n;
[ simplify;
- auto
+ autobatch
| simplify;
clear x;
clear H;
[ simplify;
elim m;
[ simplify;
- auto
+ autobatch
| simplify;
rewrite < H;
- auto
+ autobatch
]
| simplify;
rewrite < H;
clear H;
elim n;
[ simplify;
- auto
+ autobatch
| simplify;
- auto
+ autobatch
]
]
].
theorem derivative_power: ∀n:nat. D[x \sup n] = n·x \sup (pred n).
assume n:nat.
- we proceed by induction on n to prove
- (D[x \sup n] = n · x \sup (pred n)).
+ (*we proceed by induction on n to prove
+ (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 _
suppose (0=m) (m_zero). by _ done.
conclude
(D[x \sup (1+m)])
- = (D[x · x \sup m]) by _.
- = (D[x] · x \sup m + x · D[x \sup m]) by _.
- = (x \sup m + x · (m · x \sup (pred m))) by _.
-clear H.
- = (x \sup m + m · (x \sup (1 + pred m))) by _.
- = (x \sup m + m · x \sup m) by _.
- = ((1+m) · x \sup m) by _ (timeout=30)
+ = (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 + 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
done.
qed.
theorem derivative_power': ∀n:nat. D[x \sup (1+n)] = (1+n) · x \sup n.
assume n:nat.
- we proceed by induction on n to prove
- (D[x \sup (1+n)] = (1+n) · x \sup n).
+ (*we proceed by induction on n to prove
+ (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 _
(D[x \sup (2+m)] = (2+m) · x \sup (1+m)).
conclude
(D[x \sup (2+m)])
- = (D[x \sup 1 · x \sup (1+m)]) by _.
- = (D[x \sup 1] · x \sup (1+m) + x · D[x \sup (1+m)]) by _.
- = (x \sup (1+m) + x · (costante (1+m) · x \sup m)) by _.
-clear H.
- = (x \sup (1+m) + costante (1+m) · x \sup (1+m)) by _.
- = (x \sup (1+m) · (costante (2 + m))) by _
- done.
-qed.
\ No newline at end of file
+ = (D[x · x \sup (1+m)]).
+ = (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.