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Proof step 2 from proof exin (previous step)

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

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

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

Additional information

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