Stream: Archive Mirror: Isabelle Users Mailing List

Topic: [isabelle] induction on pairs of naturals

Email Gateway (Aug 18 2022 at 18:37):

From: Christian Urban <urbanc@in.tum.de>
Hi John,

There is a really cool feature of the function package that is not widely
known and that helps tremendously with such proofs. It cannot deal with
pairs, I think, but if you formulate the induction predicate as a function
taking two arguments then you can solve your problem like this.

lemma
assumes "!!n. P 0 n"
and "!!m. P m 0"
and "!!n m. [|P n (Suc m); P (Suc n) m|] ==> P (Suc n) (Suc m)"
shows "P m n"
using assms
apply(induction_schema)
apply(pat_completeness)
apply(lexicographic_order)
done

From this you original lemma should follow. Have a look at
src/HOL/ex/Induction_Schema.thy for further examples.

I think it will also help in your case with lists.

Best wishes,
Christian

John Wickerson writes:

Hello,

I'm trying to justify the following induction principle over pairs of naturals. The principle is that if you can show Phi(0,n) for all n, and Phi(m,0) for all m, and if you also show that Phi(m,n+1) and Phi(m+1,n) imply Phi(m+1,n+1) for all m and n, then you can deduce Phi(m,n) for all m and n.

My proof plan is to show that this is an instance of wf induction, but I'm really struggling! (I previously tried to justify it using ordinary mathematical induction, but got stuck on that too.) I would very much appreciate any suggestions people may have for how I can complete this proof.

lemma pair_induction:
fixes \<Phi> :: "nat \<times> nat \<Rightarrow> bool"
assumes "\<And>n2. \<Phi>(0, n2)"
assumes "\<And>n1. \<Phi>(n1, 0)"
assumes "\<And>n1 n2. \<lbrakk>\<Phi>(Suc n1, n2); \<Phi>(n1, Suc n2)\<rbrakk> \<Longrightarrow> \<Phi>(Suc n1, Suc n2)"
shows "\<And>n1 n2. \<Phi>(n1,n2)"
proof -
fix n1 n2
let ?r = "{((m1,m2),(n1,n2)) | m1 m2 n1 n2.
(Suc m1 = n1 \<and> Suc m2 = n2) \<or> (m1 = n1 \<and> Suc m2 = n2)
\<or> (Suc m1 = n1 \<and> m2 = n2)}"
have "wf ?r"
and "\<And>p. (\<And>q. (q, p) \<in> ?r \<Longrightarrow> \<Phi> q) \<Longrightarrow> \<Phi> p" sorry
from wf_induct_rule[OF this]
show "\<Phi>(n1,n2)" sorry
qed

Many thanks!

john

Last updated: Nov 21 2024 at 12:39 UTC