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Simplify, SetDelayed and Condition ... again
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52476] Simplify, SetDelayed and Condition ... again
*From*: alvaro diaz <alvarodiazfalconi at yahoo.com>
*Date*: Sun, 28 Nov 2004 01:07:19 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
This is True
>ClearAll[f,x]
>Assuming[x>0,Simplify[3x>0]]
--- True
Now define a function f[x] that f[x]:=3x/;x>0
1st try
>ClearAll[f]
>f[x_]:=3x/;x>0
>FullSimplify[f[x]>0,x>0]
>Assuming[x>0,FullSimplify[f[x]>0]]
--- f[x]>0
--- f[x]>0
--- f[x]>0
2nd try
>ClearAll[f]
>f[x_]:=3x/;Assuming[$Assumptions,x>0]
>FullSimplify[f[x]>0,x>0]
>Assuming[x>0,FullSimplify[f[x]>0]]
--- f[x]>0
--- f[x]>0
3st try
>ClearAll[f]
>f[x_]:=3x/;FullSimplify[x>0]
>FullSimplify[f[x]>0,x>0] (* 1 *)
--- f[x]>0
--- f[x]>0
but ...
>Assuming[x>0,Simplify[f[x]>0]]
--- True
FullSimplify is not hold. Why do not work (* 1 *)?
>Attributes/@{Condition,Function,FullSimplify}
---{{HoldAll,Protected},{HoldAll,Protected},{Protected}}
And the really question ... I need to modify Condition
to work better? Or there are another way without
FullSimplify in Condition?
For example, define a norm that
>norm[z_]:=Abs[z]/;FullSimplify[Element[z,Complexes]]
then
>Assuming[Element[z,Complexes],Simplify@norm[z]]
--- Abs[z]
but fail
>FullSimplify[norm[z],Element[z,Complexes]]
--- norm[z]
Alvaro.
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