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Proof step 2 from proof size-del (previous step)

a ∈ st.keys, # st.keys ≤ # (st -- a).keys ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
st.keys = ∅, a ∈ st, 0 ≤ 0 - 1 ⊦ a < st.min

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;% ⊦ # ∅ = 0; (s)%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic%leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat% ⊦ a ∈ s → #(s -- a) = # s - 1; (s)