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Proof step 22 from proof size-exgr (previous step)

(# s + 1) + 1 < # s0, a ∈ s0, ∀ s1. # s + 1 < # s1 → (∃ a0. a0 ∈ s1 ∧ a0 ≠ a1 ∧ ¬ a0 ∈ s) ⊦ a = a1, a1 ∈ s, a ∈ s, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ a0 ≠ a1 ∧ ¬ a0 ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 24:
(# s + 1) + 1 < # s0, ∀ s1. # s + 1 < # s1 → (∃ a0. a0 ∈ s1 ∧ a0 ≠ a1 ∧ ¬ a0 ∈ s), ∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1 ⊦ a = a1, a ∈ s, a1 ∈ s, ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ a0 ≠ a1 ∧ ¬ a0 ∈ s

Rule argument

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

Additional information

Used terms: s0, a