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Proof step 5 from proof subset-put-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ st[a, d] → a ∈ st0 ∧ d = st0[a], ∀ a0. a0 ∈ st[a, d] → a0 ∈ st0 ∧ st[a, d][a0] = st0[a0] ⊦ (a0 ∈ st -- a → a0 ∈ st0 -- a ∧ (st -- a)[a0] = (st0 -- a)[a0]) ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;