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Proof step 5 from proof min-delmin (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0
Proof step 7:
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

Rule argument

spec: oset inst: name: min-subset seq: ⊦ s ≠ ∅ ∧ s ⊆ s0 → s.min ∈ s0 subst: [s0, s] → [s, s -- s.min] flags: rewrite, current

Additional information

Used simplifier rules: del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union%sub-del-02: ⊦ s -- 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 -- s.min