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Proof step 12 from proof size-two (previous step)

# s0 = 1 → (∃ a. s0 = ∅ ++ a), 1 = # s0 ⊦ a ∈ s0, ∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ a = a0, ∃ a1, b. a1 ≠ b ∧ ∅ ++ a0 ++ a = ∅ ++ a1 ++ b

Rule argument

Additional information

Used simplifier rules: one-size: ⊦ s = ∅ ++ a → # s = 1;%size-one-01: ⊦ #(∅ ++ a) = 1;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;