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Proof step 14 from proof min-delmin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(s -- s.min).min = s.min, s.min ∈ s, s.max ∈ s ⊦ # s = 0, # s = 1, s = ∅, s = ∅ ++ s.min, s.max < s.min

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a < a; (s)%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem% ⊦ s -- a ⊆ s; (s)% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)