*To*: cl-isabelle-users at lists.cam.ac.uk*Subject*: Re: [isabelle] Discharging one (trivial) case without nesting the proof*From*: Christian Sternagel <c.sternagel at gmail.com>*Date*: Sun, 19 May 2013 20:18:34 +0900*In-reply-to*: <1368957723.4087.29.camel@kirk>*References*: <1368957723.4087.29.camel@kirk>*User-agent*: Mozilla/5.0 (X11; Linux x86_64; rv:17.0) Gecko/20130402 Thunderbird/17.0.5

On 05/19/2013 07:02 PM, Joachim Breitner wrote:

lemma "Q S → P S" proof assume "Q S" show "P S" proof(cases "S = {}") assume True thus "P S" by simp next assume False then obtain "x ∈ S" by auto .. show "P S" qed qed

Instead, you could write: lemma "Q S --> P S" proof assume "Q S" show "P S" proof (cases "S = {}") assume "S ~= {}" then obtain "x : S" by auto ... show "P S" qed simp qed which looks (I guess) very similar to what you suggested:

lemma "Q S → P S" proof assume "Q S" trivially "S ~= {}" proof assume "~ ~ S = {}" show "P S" by simp qed then obtain "x ∈ S" by auto .. show "P S" qed

Maybe I do not get your main point? cheers chris

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