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Proof step 3 from proof exputin (previous step)

a ∈ st[a, d] ↔ (∃ d0, st0. st[a, d] = st0[a, d0] ∧ ¬ a ∈ st0) ⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
st[a, d] = st0[a, d0] ⊦ a ∈ st0, ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;