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Proof step 3 from proof in-index-in-union (previous step)

n < m, adr ∈ asar[n] ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ union(asar, 0, m)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n < m0 +1, adr ∈ asar[n], ∀ n, adr, asar. n < m0 ∧ adr ∈ asar[n] → # asar[n] = 0 ∨ asar[n] = ∅ ∨ adr ∈ union(asar, 0, m0) ⊦ # asar[n] = 0, asar[n] = ∅, adr ∈ union(asar, 0, m0 +1)

Rule argument

indvar: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ ¬ m < 0; (s)