From: Randy Pollack <rpollack@inf.ed.ac.uk>
I have mutually defined inductive relations on a datatype:
datatype preClam =
pcVar nat "nat list"
| pcLam preClam
inductive weak :: "nat \<Rightarrow> preClam \<Rightarrow> preClam
\<Rightarrow> bool"
and split :: "preClam \<Rightarrow> (nat * preClam) \<Rightarrow> bool"
where
wkVar[intro!]: "weak n (pcVar m gam) (pcVar m (Cons n gam))"
| wkLam[intro]: "\<lbrakk>split wmM (m,M); weak n M wnM; weak m wnM
wmnM\<rbrakk>\<Longrightarrow>
weak n (pcLam wmM) (pcLam wmnM)"
| split[intro!]: "weak n N wnN \<Longrightarrow> split wnN (n,N)"
I want to prove a theorem by simultaneous induction:
lemma
shows weak_pcPN:"weak m M wmM \<Longrightarrow> P"
and split_pcPN:"\<lbrakk>split wnN nN; nN = (n,N)\<rbrakk>
\<Longrightarrow> Q"
I want to generalize n and N which appear in the second part of the
lemma. (This is standard, as n and N were only introduced to make the
statement of the lemma suitable for induction.) So I expect to write
something like
proof (induct arbitrary: n N rule: weak_split.inducts)
This attempt applies the induction rule but doesn't generalize n and
N. The relevant fact introduced by this induction is
(m, M) = (n, N) \<Longrightarrow> Q
which is of no use at all since n and N are not generalized.
How can this proof be done?
Thanks for any help.
Randy
From: Tobias Nipkow <nipkow@in.tum.de>
Hi Randy,
I think you need to give arbitrary a list of lists, one for each
inductive predicate, separated by "and". In this case the first list
should probably be empty:
proof (induct arbitrary: and n N rule: weak_split.inducts)
Tobias
Last updated: Nov 21 2024 at 12:39 UTC