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Proof step 8 from proof And-discard (previous step)

a ∈ s, ∀ s0, P. # s0 < # s → true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P ⊦ # s = 0, (λ a. P) ∧_ s = P

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 10:
∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0, ∀ s0, P. # s0 < # s → true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P ⊦ # s = 0, (λ a. P) ∧_ s = P

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 terms: s, a