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Proof step 1 from proof size-del (conclusion)

⊦ a ∈ st.keys → ¬ # st.keys ≤ # (st -- a).keys

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-01: ⊦ a ∈ st.keys → (st -- a).keys = st.keys -- a;%Keys: ⊦ a ∈ st.keys ↔ a ∈ st;%Min-min: ⊦ a ∈ st → ¬ a < st.min;%inf: ⊦ a ∈ st → a ∈ st.keys;%notempty: ⊦ a ∈ st.keys → st.keys ≠ ∅;%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%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%del-size: ⊦ a ∈ s → #(s -- a) = # s - 1; from specification set-union