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Proof step 3 from proof in-01 (previous step)

∀ s0, a. # s0 < # s → true → (a ∈ set2list(s0) ↔ a ∈ s0) ⊦ a ∈ set2list(s) ↔ a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s ≠ ∅ → set2list(s) = s.min + set2list(s.butmin)
Proof step 5:
s ≠ ∅, ∀ s0, a. # s0 < # s → true → (a ∈ set2list(s0) ↔ a ∈ s0) ⊦ a ∈ s.min + set2list(s.butmin) ↔ a ∈ s

Rule argument

name: set2list-rec rotate: F subst: [s] → [s]

Additional information

Used simplifier rules: set2list-empty: ⊦ set2list(∅) = [];% ⊦ ¬ a ∈ [];% ⊦ ¬ a ∈ ∅; Used terms: s