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Proof step 1 from proof min-in (conclusion)

⊦ min(ints) ∈ ints ↔ ints ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ints = [], min(ints) ∈ ints

Rule argument

Additional information

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