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Proof step 1 from proof butmin-size (conclusion)

⊦ st ≠ ∅ → ¬ # st ≤ # st.butmin

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ st.butmin = st -- st.min;%min-def: ⊦ st ≠ ∅ → st.min = st.keys.min;%Min-in: ⊦ st ≠ ∅ → st.min ∈ st;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%empty: ⊦ # st = 0 ↔ st = ∅; from specification store%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%pred-03: ⊦ n -1 = n - 1; from specification nat%size: ⊦ a ∈ st → #(st -- a) = (# st)-1; from specification store