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Proof step 2 from proof bmin-bneq (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 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: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%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%one-min: ⊦ s = ∅ ++ a → s -- a = ∅; from specification set-union%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic%in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic Used terms: s -- s.min