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Proof step 5 from proof min-delmin (previous step)

s.max ∈ s, s.min ∈ s ⊦ # s = 1, # s = 0, s = ∅, s.max < (s -- s.min).min, s.min < (s -- s.min).min, s.max < s.min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
s.min ∈ s, s.max ∈ s ⊦ # s = 0, # s = 1, s = ∅, s.max < s.min, s.min < (s -- s.min).min, s.max < (s -- s.min).min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

Proof step 5 from proof min-delmin (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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