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Proof step 1 from proof bmin-size (conclusion)

⊦ # s.butmin < # s ↔ s ≠ ∅

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;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n ≤ m ↔ n = 0; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ a ∈ s → #(s -- a) = # s - 1; (s)%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ¬ n < 0; (s)% ⊦ # ∅ = 0; (s)