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Proof step 10 from proof butmin-rewput (previous step)

a0 = st[a, d].min ⊦ a = st[a, d].min, a0 = a, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st -- st.min) ∧ (st[a, d] -- st[a, d].min)[a0] = (st -- st.min)[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
st.min < st[a, d].min, st[a, d].min ∈ st ⊦ st[a, d].min = st.min, a = st[a, d].min

Rule argument

Additional information

Used simplifier rules: Min-min: ⊦ a ∈ st → ¬ a < st.min;%inf: ⊦ a ∈ st → a ∈ st.keys;%Keys: ⊦ a ∈ st.keys ↔ a ∈ st;% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)% ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (ws)% ⊦ ¬ a ∈ st → st[a] = (∅)[a]; (s)% ⊦ (st -- a)[a] = (∅)[a]; (s)% ⊦ ¬ a ∈ st -- a; (s)