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Proof step 2 from proof max-in-not-ls (previous step)

i ∈ ints, max(ints) < i ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ ints ≠ [] → max(ints) ∈ ints ∧ (∀ i. i ∈ ints → i ≤ max(ints))

Rule argument

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

Additional information

Used simplifier rules: max-in-01: ⊦ max(ints) ∈ ints ↔ ints ≠ [];%max-empty: ⊦ max([]) = 0;%max-is-max: ⊦ ints ≠ [] ∧ i ∈ ints → ¬ max(ints) < i;%in-empty: ⊦ ¬ i ∈ []; from specification list[intlista] Used terms: ints