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Proof step 13 from proof min-delmin (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ # ∅ = 0; (s)