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Proof step 7 from proof size-one-02 (previous step)

a0 ∈ s0, #(s0 ++ a0) = 1, a ∈ s0, b = a0 ∨ b ∈ s0, ∀ b, a. # s0 = 1 ∧ b ∈ s0 ∧ a ∈ s0 → a = b ⊦ a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# s0 = 1, a0 ∈ s0, a ∈ s0, b ∈ s0, ∀ b0, a1. b0 ∈ s0 ∧ a1 ∈ s0 → a1 = b0 ⊦ a = b

Rule argument

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;