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Proof step 8 from proof bm-shift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ≠ s.max ∧ s ≠ ∅ → (s -- a).max = s.max
Proof step 10:
s.max ∈ s, s.min ∈ s ⊦ # s = 1, # s = 0, s -- s.min -- (s -- s.min).max = s -- s.max -- (s -- s.max).min, s = ∅ ++ s.max, s = ∅ ++ s.min, s = ∅, s.max < s.min, s.min ≠ s.max ∧ s ≠ ∅

Rule argument

spec: <toplevel> name: max-del-01 seq: ⊦ a ≠ s.max ∧ s ≠ ∅ → (s -- a).max = s.max subst: [a, s] → [s.min, s] flags: <all paths>

Additional information

Used simplifier rules: min-del: ⊦ s ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min;%bmin-del-02: ⊦ s ≠ ∅ → s -- a -- s.min = s.butmin -- a;%Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min; Used terms: s.min, s