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Proof step 12 from proof subset-put-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
a0 ∈ st, (∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a]) ∧ ¬ a ∈ st ∨ a ∈ st ∧ st[a] = d ∧ (∀ a0. a0 ∈ st -- a → a0 ∈ st0 -- a ∧ st[a0] = st0[a0]) ⊦ a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0]

Proof step 12 from proof subset-put-01 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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