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Proof step 5 from proof max-rec (previous step)

∀ i. i ∈ ints → i ≤ max(ints) ⊦ ints = [], max(i ' + ints) = max(i, max(ints))

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
⊦ ints ≠ [] → max(ints) ∈ ints ∧ (∀ i. i ∈ ints → i ≤ max(ints))
Proof step 7:
i ' + ints ≠ [] → max(i ' + ints) ∈ i ' + ints ∧ (∀ i0. i0 ∈ i ' + ints → i0 ≤ max(i ' + ints)), ∀ i. i ∈ ints → i ≤ max(ints) ⊦ ints = [], max(i ' + ints) = max(i, max(ints))

Rule argument

spec: inst: name: max-def seq: ⊦ ints ≠ [] → max(ints) ∈ ints ∧ (∀ i. i ∈ ints → i ≤ max(ints)) subst: [ints] → [i ' + ints] flags: current, []

Additional information

Used terms: i ' + ints