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Proof step 9 from proof size-two (previous step)

# s = 2, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
1 = # s0 ⊦ a ∈ s0, ∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-num-04: n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; from specification nat