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Proof step 3 from proof bmax-one-01 (previous step)

# s = 1, s.butmax.min ∈ s.butmax, s.butmax.max ∈ s.butmax ⊦ # s.butmax = 0, s.butmax = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s ≠ ∅ → s.butmax = s -- s.max

Rule argument

spec: <toplevel> name: Butmax seq: ⊦ s ≠ ∅ → s.butmax = s -- s.max subst: [s] → [s] flags: <all paths>

Additional information

Used simplifier rules: max-one-01: ⊦ # s = 1 → (s = ∅ ++ s.max ↔ true);% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)% ⊦ # ∅ = 0; (s) Used terms: s