Re: [isabelle] Nat theory proofs



Sorry, the 2 had type 'a. 
Tim 



________________________________
From: TIMOTHY KREMANN <twksoa262 at verizon.net>
To: Isabelle List <cl-isabelle-users at lists.cam.ac.uk>
Sent: Wednesday, December 24, 2008 9:38:53 AM
Subject: Nat theory proofs


I am trying to prove:

lemma nataba: "\<forall> a b. (b::nat) < a --> a * a - (2 * b * a - b * b) = 
                              a * a - 2 * a * b + b * b"

But Isabelle returns this text when I enter the above:



proof (prove): step 0

goal (1 subgoal):
 1.  \<forall> a b.
       b < a -->
       a * a - (2 * b * a - b * b) =
       a * a - 2 * a * b + b * b

Counterexample found:

a = Suc (Suc (Suc 0))
b = Suc (Suc 0)

Can someone explain to me how 1 = 1 is a counterexample?

Thanks,
Tim 




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