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Proof step 7 from proof min-delmin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
(s -- s.min).min = s.min, s.min ∈ s, s.max ∈ s ⊦ s = ∅, # s = 0, # s = 1, s = ∅ ++ s.min

Rule argument

Additional information

Used simplifier rules: irreflexivity: ⊦ ¬ a < a; from specification oelem%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem%sub-del-02: ⊦ s -- a ⊆ s; from specification set-union%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union