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Proof step 7 from proof subset-put (previous step)

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

Additional information