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Proof step 18 from proof subset-put-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information