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Proof step 3 from proof subset-put-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0] ⊦ a ∈ st ∧ st[a] = d ∧ (a0 ∈ st -- a → a0 ∈ st0 -- a ∧ (st -- a)[a0] = (st0 -- a)[a0]), (a1 ∈ st → a1 ∈ st0 ∧ st[a1] = st0[a1]) ∧ ¬ a ∈ st

Proof step 3 from proof subset-put-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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