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Proof step 3 from proof And-unfold (previous step)

a ∈ s ⊦ # s = 0, s = ∅, aP ∧_ s = (aP(a) ∧ aP ∧_ (s -- a))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)

Rule argument

spec: set-basic name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s, a] flags: <all paths>

Additional information

Used simplifier rules: And-ins: ⊦ ¬ a ∈ s → aP ∧_ (s ++ a) = (aP(a) ∧ aP ∧_ s);% ⊦ ¬ a ∈ s → (s ++ a) -- a = s; (s) Used terms: s, a