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Proof step 5 from proof butmin-put-01 (previous step)

a = st[a, d0].min ⊦ a ∈ st, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st) ∧ (st[a, d] -- st[a, d].min)[a0] = st[a0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ≠ b → (st -- b)[a] = st[a]
Proof step 7:
¬ a0 ≠ st[a, d].min, a = st[a, d0].min ⊦ a ∈ st, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st) ∧ (st[a, d] -- st[a, d].min)[a0] = st[a0]
Proof step 9:
a0 ≠ st[a, d].min, a = st[a, d0].min ⊦ a ∈ st, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st) ∧ st[a, d][a0] = st[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]