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Proof step 2 from proof bmax-shift (previous step)

⊦ s -- a = ∅, (s -- a).butmax = s.butmax -- a, a = s.max

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmax: ⊦ s ≠ ∅ → s.butmax = s -- s.max;%max-del-01: ⊦ a ≠ s.max ∧ s ≠ ∅ → (s -- a).max = s.max;% ⊦ s -- b -- a = s -- a -- b ↔ true; (s)% ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; (ws)