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Proof step 5 from proof butmin-rewput (previous step)

⊦ a = st[a, d].min, (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 weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
⊦ a0 = a, a = st[a, d].min, (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]

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: insin: ⊦ a ∈ st[a, d]; from specification store%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; from specification store%Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a]; from specification store%At-same: ⊦ st[a, d][a] = d; from specification store Used formulas: a0 = a