# Re: [isabelle] nat properties

```kuecueck,

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Here is a proof of your lemma. Unfortunately it includes certan non trivial steps, whith whom automatic tactics do not deal.
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The key here is to give a witness divisor for "nat a dvd nat b" which is "nat k", where k is the witness divisor for "a dvd b".
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Amine.

lemma assumes ap:"(0::int)<a" and bp:"0<b" and d: "a dvd b"
shows "(nat a) dvd (nat b)"
using ap bp d
proof-
from d have " EX k. b = a*k" unfolding dvd_def .
then obtain k where bak: "b= a*k" by blast
with ap bp have kp: "k >= 0"
using mult_less_cancel_right_disj[where a="0" and c="k" and b="a"]
by auto
from ap bp have ap': "a >= 0" and bp':"b >= 0" by simp+
from mult_nonneg_nonneg[OF ap' kp] have akp:"a*k >= 0" .
from nat_mult_distrib[OF ap', where z'="k", symmetric] bak
eq_nat_nat_iff[OF bp' akp] have "nat b = nat a * nat k" by simp
thus ?thesis by auto
qed

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```Hallo

i need some properties of the nat.

i couldn't proof the lemma below.
[|(0::int)<a; 0<b; a dvd b|]==> (nat a) dvd (nat b)

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is there any theory which include such properties or is there a solution to prove this lemma?
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thanks
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