How can I prove irreflexivity of an inductively defined relation in Isabelle?

Question

Consider as an example the following definition of inequality of natural numbers in Isabelle:

inductive unequal :: "nat ⇒ nat ⇒ bool" where
  zero_suc: "unequal 0 (Suc _)" |
  suc_zero: "unequal (Suc _) 0" |
  suc_suc:  "unequal n m ⟹ unequal (Suc n) (Suc m)"

I want to prove irreflexivity of unequal, that is, ¬ unequal n n. For illustration purposes let me first prove the contrived lemma ¬ unequal (n + m) (n + m):

lemma "¬ unequal (n + m) (n + m)"
proof
  assume "unequal (n + m) (n + m)"
  then show False
  proof (induction "n + m" "n + m" arbitrary: n m)
    case zero_suc
    then show False by simp
  next
    case suc_zero
    then show False by simp
  next
    case suc_suc
    then show False by presburger
  qed
qed

In the first two cases, False must be deduced from the assumptions 0 = n + m and Suc _ = n + m, which is trivial.

I would expect that the proof of ¬ unequal n n can be done in an analogous way, that is, according to the following pattern:

lemma "¬ unequal n n"
proof
  assume "unequal n n"
  then show False
  proof (induction n n arbitrary: n)
    case zero_suc
    then show False sorry
  next
    case suc_zero
    then show False sorry
  next
    case suc_suc
    then show False sorry
  qed
qed

In particular, I would expect that in the first two cases, I get the assumptions 0 = n and Suc _ = n. However, I get no assumptions at all, meaning that I am asked to prove False from nothing. Why is this and how can I conduct the proof of inequality?

Answer 1

The induction method handles variable instantiations and non-variable instantiations differently. A non-variable instantiation t is a shorthand for x ≡ t where x is a fresh variable. As a result, induction is done on x, and the context additionally contains the definition x ≡ t.

Therefore, (induction "n + m" "n + m" arbitrary: n m) in the first proof is equivalent to (induction k ≡ "n + m" l ≡ "n + m" arbitrary: n m) with the effect described above. To get this effect for the second proof, you have to replace (induction n n arbitrary: n) with (induction k ≡ n l ≡ n arbitrary: n). The assumptions will actually become so simple that the pre-simplifier, which is run by the induction method, can derive False from them. As a result, there will be no cases left to prove, and you can replace the whole inner proof–qed block with by (induction k ≡ n l ≡ n arbitrary: n).