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Proof step 7 from proof size-one (previous step)

# s = 1, a ∈ s ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 9:
# s = 1, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Rule argument

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

Additional information

Used terms: s, a