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Proof step 16 from proof subset-put-01 (previous step)

(∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a]) ∧ ¬ a ∈ st, 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