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Proof step 6 from proof size-one-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#(s0 ++ a0) = 1, a ∈ s0, ∀ a, b. # s0 = 1 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ a = b

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);