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Proof step 1 from proof exputin (conclusion)

⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ a ∈ st ↔ (∃ d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 3:
a ∈ st[a, d] ↔ (∃ d0, st0. st[a, d] = st0[a, d0] ∧ ¬ a ∈ st0) ⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Rule argument

spec: store inst: name: exin seq: ⊦ a ∈ st ↔ (∃ d, st0. st = st0[a, d] ∧ ¬ a ∈ st0) subst: [a, st] → [a, st[a, d]] flags: current, []

Additional information

Used terms: a, st[a, d]