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Proof step 4 from proof bmin-del-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
s.min ∈ s, s.max ∈ s ⊦ # s = 0, a = s.min, s = ∅, (s -- a).butmin = s -- s.min -- a

Rule argument

Additional information

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