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Proof step 1 from proof bmax-shift (conclusion)

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

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;%delsw-02: ⊦ s -- b -- a = s -- a -- b ↔ true; from specification set-union%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union