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Proof step 2 from proof size-three-01 (previous step)

# s < 3, c ∈ s, b ∈ s, a ∈ s ⊦ a = b, a = c, b = c

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 4:
# s < 3, c ∈ s, b ∈ s, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ a = b, a = c, b = c

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