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Proof step 8 from proof size-two-03 (previous step)

s0 = s ++ a2, # s0 ≠ 1 → a0 ≠ b ∧ a0 ∈ s0 ∧ b ∈ s0 ⊦ # s0 = 0, s0 = ∅, s0 = s, a ∈ s0, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 ∈ s0) ∧ (b0 = a ∨ b0 ∈ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# s ≠ 0 → a0 ≠ b ∧ (a0 = a2 ∨ a0 ∈ s) ∧ (b = a2 ∨ b ∈ s) ⊦ a = a2, a2 ∈ s, a ∈ s, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 = a2 ∨ a1 ∈ s) ∧ (b0 = a ∨ b0 = a2 ∨ b0 ∈ s)

Rule argument

Additional information

Used simplifier rules: eq-03: ⊦ s ++ a = s ↔ a ∈ s;%ne: ⊦ ∅ ≠ s ++ a;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;% ⊦ 1 = m + 1 ↔ m = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)