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Proof step 8 from proof min-delmin (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 9:
⊦ s ≠ ∅ → s.min ∈ s

Rule argument

spec: oset inst: name: Min-in seq: ⊦ s ≠ ∅ → s.min ∈ s subst: [s] → [s -- s.min] flags: current, []

Additional information

Used simplifier rules: del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s); from specification set-union Used terms: s -- s.min