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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ≠ b → (st -- b)[a] = st[a]
Proof step 9:
¬ a0 ≠ st[a, d].min ⊦ a0 = a, a = st[a, d].min, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st -- st.min) ∧ (st[a, d] -- st[a, d].min)[a0] = (st -- st.min)[a0]
Proof step 14:
a0 ≠ st[a, d].min ⊦ a0 = a, a = st[a, d].min, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st -- st.min) ∧ st[a, d][a0] = (st -- st.min)[a0]

Rule argument

spec: store inst: name: Del-at seq: ⊦ a ≠ b → (st -- b)[a] = st[a] subst: [a, b, st] → [a0, st[a, d].min, st[a, d]] flags: rewrite

Additional information

Used terms: a0, st[a, d].min, st[a, d]