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Proof step 7 from proof butmin-rewput (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ 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]

Rule argument

Additional information

Used simplifier rules: ⊦ a ≠ b → st[b, d][a] = st[a]; (s)% ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); (s)