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Proof step 2 from proof bmin-eq-01 (previous step)

s.butmin = s0.butmin, s.min = s0.min ⊦ s = ∅, s = s0, s0 = ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;% ⊦ a ∈ s ∧ a ∈ s0 → (s -- a = s0 -- a ↔ s = s0); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic