[isabelle] can't hack induction



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Ok, so I'm trying to get started. So far I've only been doing things
that can be proven by auto. Now I'm trying induction and things aren't
working. I'm trying to prove that "successor" can be "factored" out of
add:

datatype n =
  Z
| S n

primrec add :: "n => n => n"
where
  "(add x Z) = x"
| "(add x (S y)) = (add (S x) y)"

lemma desired: "!!y. !!x. ((add x (S y)) = (S (add x y)))"

I can write out the steps of the induction I'm trying to do:

!!x. add x sz = add sx z = s (add x z) # this is the base case y=Z
!!x. add x ssz = add sx sz = s (add sx z) = s (add x sz)
!!x. add x sssz = add sx ssz = s (add sx sz) = s (add x ssz)

So the general idea is to do a "forward step" (where an s is shifted
left), an "inductive step" and a "backward step" (where an s is
shifted right). The base case and the inductive step can be proven
with auto:

lemma bascase: "(!!x. (add x (S Z)) = (S (add x Z)))"
apply(auto)
done

lemma indstep [simp]: "(!!x. (add x (S y)) = (S (add x y))) ==> (!!x.
(add x (S (S y))) = (S (S (add x y))))"
apply(auto)
done

However, if I try to apply induct_tac on the desired lemma:

lemma desired: "!!y. !!x. ((add x (S y)) = (S (add x y)))"
apply(induct_tac y)
apply(auto)

I get:

proof (prove): step 2

goal (1 subgoal):
 1. !!x n. add (S x) n = S (add x n) ==> add (S (S x)) n = S (S (add x n))

This is not what I want. My induction is based on !!y (!!x p(x,y)) ->
(!!x p'(x,y)) but what it is asking me to prove is !!y !!x (p(x,y) ->
p'(x,y)). How do I get back on track?

-Andrei





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