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Proof step 11 from proof min-delmin (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 12:
#(∅ ++ s.min) = # ∅ + 1, s.max = s.min, s = ∅ ++ s.min, s.min ∈ s ⊦ # s = 1, # s = 0, s = ∅, s.min < (s -- s.min).min, (s -- s.min).min ≠ s.min ∧ (s -- s.min).min ∈ s

Rule argument

pos: left 4

Additional information

Used simplifier rules: ⊦ a ∈ s → s ++ a = s; (s)