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Proof step 4 from proof min-in-not-ls (previous step)

ints ≠ [] → min(ints) ∈ ints ∧ (∀ i. i ∈ ints → min(ints) ≤ i), i ∈ ints, i < min(ints) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i ∈ ints, i < min(ints), ∀ i. i ∈ ints → min(ints) ≤ i ⊦ ints = []

Rule argument

Additional information

Used simplifier rules: min-in: ⊦ min(ints) ∈ ints ↔ ints ≠ [];%min-empty: ⊦ min([]) = 0;%in-empty: ⊦ ¬ i ∈ []; from specification list[intlista]