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Proof step 9 from proof subset-put-02 (previous step)

st0[a] = d, a ∈ st0, ∀ a0. a0 ∈ st -- a → a0 ∈ st0 -- a ∧ (st -- a)[a0] = (st0 -- a)[a0] ⊦ ∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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