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Proof step 3 from proof butmin-ins-01 (previous step)

a < s.min, s.min ∈ s, s.max ∈ s ⊦ # s = 0, (s ++ a).butmin = s, s = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s ≠ ∅ → s.butmin = s -- s.min

Rule argument

spec: <toplevel> name: Butmin seq: ⊦ s ≠ ∅ → s.butmin = s -- s.min subst: [s] → [s ++ a] flags: <all paths>

Additional information

Used simplifier rules: min-insert: ⊦ a < s.min ∧ s ≠ ∅ → (s ++ a).min = a;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;% ⊦ (s ++ a) -- a = s -- a; (s)% ⊦ s -- a = s ↔ ¬ a ∈ s; (s)% ⊦ a ∈ s → s ++ a = s; (s)% ⊦ ∅ ≠ s ++ a; (s) Used terms: s ++ a